1D

KINEMATICS PHYSICS - I A

AKASH MULTIMEDIA 124

4.1 INTRODUCTIONThe study of mot ion of bodies without

consider ing the cause of mot ion is cal ledKinematics. In the universe every object moves,though some objects appear to be stationary. Theterms motion and rest are relative. We begin thischapter with motion along a straight line, i.e. motionin one dimension. Later, we look into motion in aplane, i.e. two-dimensional motion.4.2 REST, MOTION

If a particles position does not change eitherwith respect to a fixed point and with respect to time,it is said to be at rest.

If a particles position is continuously changingwith respect to a fixed point and with respect to time,it is said to be in motion.4.3 MOTION IN A STRAIGHT LINE

When a particle is in motion, if the line joiningthe successive positions is a straight line then it issaid to be in straight line motion. Here, the referencepoint about which the position of the particle islocated is called origin.4.4 DISTANCE AND DISPLACEMENT

Length of actual path between initial andfinal positions is called distance.

The shor test straight line distance directedfrom initial position to final position ir respectiveof the path is called displacement.

The distance and displacement are twodifferent quantities. Distance has just a magnitude

(numerical value) and no direction. whereas, thedisplacement has magnitude as well as direction.Hence, distance is scalar and displacement is vector.

Application 4.1: Consider a particle movingfrom A to B along a curve as shown.

A BThe distance travelled is equal to the length of

the curve AB, whereas the magnitude of thedisplacement is equal to the length of the straightline AB.

Application :4.2 A B

C

3m

5m 4m

If a person walks from A to B and then from Bto C as shown,

Distance travelled = AB + BC= 7mDisplacement = AC = 5m

Application 4.3 : A particle moves over anarc PQ , of a circle of radius R, subtending an angle at the centre .

P Q

R R

O

a) distance travelled = arc PQ = R

b) displacement= straight line PQ = 2R sin2

KINEMATICS

h Straight line motionh Motion under gravity

CHAPTER

4Alber t Einstein

Alber t Einstein (14 March 1879 18 April 1955) was a German-born theo-retical physicist who discovered the theory of general relativity, effecting a revolu-tion in physics. For this achievement, Einstein is often regarded as the father ofmodern physics. He received the 1921 Nobel Prize in Physics "for his services totheoretical physics, and especially for his discovery of the law of the photoelectriceffect".

He continued to deal with problems of statistical mechanics and quantumtheory, which led to his explanations of particle theory and the motion of molecules.He also investigated the thermal properties of light which laid the foundation of thephoton theory of light. In 1917, Einstein applied the general theory of relativity tomodel the structure of the universe as a whole.

h Oblique projectileh Horizontal projectile

PHYSICS - I A KINEMATICS


Note 4.1 : If a particle starts from a point andreaches the same point at the end of its journey, thendisplacment is zero. However distance covered isnot zero. Therefore, a particle can travel somedistance without displacement.

Note 4.2 : For a particle in motion,the magnitudeof displacement cannot exceed the distancedisplacem ent distance (for curved motion).

displacem ent distance (for staright linemotion).4.5 SPEED

The distance travelled by a body in unit timeis called its speed.

It is a scalar quantity.CGS unit is cms1 and SI unit is ms1.

4.6 VELOCITYThe displacement of a body in unit time is

called its Velocity.It is a vector quantity.CGS unit is cms1 SI unit is ms1.

4.7 UNIFORM MOTIONIf a particle moving along a straight line (say x

axis) travels equal distances in equal intervals of timewe say the particle is in uniform motion. The motionis expressed by an equation of the form.

x = vt + bwhere x is the position coordinate of the particle,

t is the time, v and b are certain constants. In thisequation position is a linear function of time. Hencethe slope of the position - time graph is a straightline, the slope of which is a constant( v) and is equalto constant velocity or uniform velocity of particle.

t

x

b

o

Position time graph of an object in uniformmotion.

When t = 0, the above equation can be written as x0 = bWhere x0 indicates the initial position of the

particle from the origin.

4.8 NON UNIFORM MOTION ORVARIABLE MOTION If a particle moving along a straight line travels

unequal distances in equal intervals of time or equaldistances in unequal intervals of time, we say thatthe particle is in non-uniform motion.

Ex : i) Motion of a freely falling body.The equation of the motion is 2

1y gt2

.The positiontime graph is a parabola with

increasing slope.

t

y

o

Position time graph of a freely falling body in non- uniform motion.

Ex : ii) Motion of a body thrown ver ticallyupwards.

The equation of motion is 21y ut gt2

The positiontime graph is a parabola.Its slope decreases from + K to zero and

thereafter from zero to K, where K is instantaneousslope.

t

y

oPosit ion t ime gr aph of a body thr own

vertically upwards.4.9 UNIFORM SPEED

If a body travels equal distances in equalintervals of time however small the intervals maybe, then it is said to be moving with uniform speed.

4.10 NON UNIFORM SPEEDIf a body travles equal distances in unequal

intervals of time or unequal distances in equal intervalsof time then it is said to be moving with non-uniformspeed.



4.11 INSTANTANEOUS SPEEDThe speed of a particle at a particular instant of

time is called its instantaneous speed.If S is the distance travelled by a particle in a

time interval t then

speed = St

If the time interval t is chosen to be very small,i.e., as t 0 , then the corresponding speed is calledinstantaneous speed.

t 0

S dSLt instantaneous speed.t dt

4.12 AVERAGE SPEEDFor a particle in motion (uniform or non-

uniform), the ratio of total distance travelled to thetotal time of motion is called average speed.

Average speed =Total distance

Total timeIf s1, s2, s3......sn are the distances travelled by a

particle in the time intervals t1, t2, t3......tn respectivelythen,

1 2 3 n

1 2 3 n

s s s ......sAverage Speedt t t .......t

Application 4.4 :A body travelling between two positions traveles

first half of the distance with speed v1 and the nexthalf of the distance with speed v2 The average speed

of total motion is 1 2

1 2

2v vv + v

Let x be the total distance between two positions.Let t1 be the time for first half and t2 be the time

for the next half of the distance

= Total distance Avg. speedTotal time

1 2

1 2

v vv + v

v v1 2

1 2

x x 2x xt t

2 2

Application 4.5:A body is travelling between two positions A,

B. It travelles from A to B with speed v1 and then

from B to A in the same path with speed v2 . The

average speed of total motion is 1 21 2

2v vv + v

Application 4.6 :A body is travelling between two positions. The

total distance is divided into n equal parts. These partsare travelled with speed v1, v2, v3.....vn respectively.The average speed of total motion is such that

v v v v1 2 3 n

n 1 1 1 1.........Average speed

Application 4.7:A body travelling between two positions traveles

with speed v1 for time t1 and then with speed v2 fortime t2. For the total motion,

= 1 1 2 21 2

v t + v tAverage Speedt + t

Application 4.8:A body travelling between two positions travels

first half of the time with speed v1 and the next halfof the time with speed v2. The average speed of total

motion is 1 2v + v

2Application 4.9 :

A body travelling between two positions travelsfor the time intervals t1, t2, t3..........tn with speeds v1,v2, v3,...........vn respectively

totaldistanceAvg. speed =total time

=1 1 2 2 3 3 n n

1 2 3 n

v t + v t + v t +............v tt + t + t +.........t

4.13 UNIFORM VELOCITYIf a body has equal displacements in equal

intervals of time however small the intervals may bethen it is said to be moving with uniform velocity.4.14 NON-UNIFORM VELOCITY

If a body has equal displacements in unequalintervals of time or unequal displacements in equalintervals of time then it is said to be moving withNonUniform Velocity.

Note : The displacement variation may be dueto change in magnitude or change in direction ofmotion or both.



4.15 INSTANTANEOUS VELOCITYThe velocity of a particle at a particular instant

of time is called its instantaneous velocity.If S

is the displacement by a particle in a timeinterval t then

SVelocity Vt

If the time interval t is chosen to be very small,i.e., as t 0 , the corresponding velocity is calledinstantaneous velocity.

t 0

S dSLt instantaneous velocityt dt

The instantaneous velocity is rate of change ofposition with time.

Note 4.3 : 2 2

1 1

s t

s t

dsv ; ds vdtdt

4.16 Average velocity :For a particle in motion (uniform or non-

uniform), the ratio of total displacement to the totaltime interval is called Average velocity.

Total displacementAverage velocity =Total time

Suppose a particle displaces from P1 to P2 in atime interval t. If 1x

is initial position vector and

2x

is final position vector then

= 2 1x xAverage velocity vt

Note 4.4 : A particle travelling between twopositions A, B travels from A to B with velocity v1and returns from B to A with velocity v2. Nowaverage velocity of total motion is zero since the netdisplacement of the particle is zero.

Note 4.5 : In variable motion, the averagevelocity depends on the interval of the time duringwhich the velocity is calculated.

Note 4.6 : In uniform motion, the averagevelocity is a constant and is same for all the intervalsof time during which the value is calculated.

Note 4.7 : The magnitude of average velocity isequal to the average speed for motion along a straightline and it is a scalar quantity.

Note 4.8 : In uniform motion, the instantaneousvelocity of a body is equal to the average velocity.4.17 ACCELERATION

If the velocity of a particle is changing as itmoves then it is said to be moving with acceleration.The acceleration measures how rapidly the velocityis changing.

Acceleration is defined as the rate of changeof velocity.

Let 1 2V ,V

be the velocities of a particle atinstances t1, t2 respectively. Now,

change in velocityAcceleration time

= 2 12 1

V - V DVat - t Dt

I t is a vector. I t is in the direction of changein velocity.

S.I. Unit is ms2, dimensional formula is [L1T2]Note 4.9 : The velocity variation may be due to

change in magnitude of velocity (speed) or changein direction of velocity. Hence acceleration may bedue to either of the above reasons or both.

eg 1 : For a car going on a straight road if thespeed is increasing, then the acceleration is due tochange in magnitude of velocity

eg 2 : For a stone whirled in a horizontal circlewith constant speed, the acceleration is due to changein direction of velocity.

eg 3 : For a stone whirled in a vertical circlewith a changing speed, the acceleration is due tochange in both magnitude and direction of velocity.

Note 4.10 : The acceleration of a moving particlemay be positive or negative. If the speed of particleis increasing with time then acceleration is positiveand if the speed is decreasing with time thenacceleration is negative.

Note 4.11 : For positive acceleration the velocityvector and acceleration vector are in the samedirection. But for negative acceleration, the velocityand acceleration vectors are opposite



4.18 DECELERATION OR RETARDATIONIf the speed is decreasing with time then

acceleration is negative.The negative acceleration is called decelera

tion or retardation.

4.19 INSTANTANEOUS ACCELERATIONThe acceleration of a particle at a particular

instant of time is called its instantaneous acceleration.

As the velocity of a particle is changing withtime, if the time interval approaches to zero (i.e., toan infinitely short time interval)

t 0

V Vt

dLtdt

Instantaneous acceleration

Note 4.12 : Instantaneous accelar ation

2 2

1 1

2 v t

2v t

dv d sa ; dv a dtdt dt

Note 4.13 : dv ds dva . ; a v.ds dt ds

ads vdv 4.20 UNIFORM ACCELERATION

If the average acceleration over any time intervalequals the instantaneous acceleration at any instantof time then the acceleration is said to be uniform orconstant. It does not vary with time. The velocityeither increases or decreases at the same ratethroughout the motion.

4.21 KINEMATICAL EQUATIONS OF MOTION OF A PARTICLE MOVING ALONG A STRAIGHT LINE WITH UNIFORM ACCELERATION :Kinematical equations are useful to solve

problems in one dimensional motion of a particlewith constant acceleration.

Consider a particle with initial position vector

ix

. Suppose it starts with initial velocity u

andmoves with uniform acceleration a. Suppose v is its

final velocity after t seconds with fx

as final positionvector.

Now the equations of motion are as follows.

1) Velocity as a function of time v u at

(or)v = u + at

2) Displacement as a function of time

2f i1S x x ut at2

(or) + 2

1S = ut at2

3) Position as a function of time

2

f i1x x ut at2

(or) 2f i1x x ut at2

4) Velocity as a function of displacementv.v u.u 2 a.s

(or) v2u2 = 2as

5) Displacement in thn second of motion

= +n1S u a n -2

(or) n

1S = u + a n -2

6) Displacement = (Average velocity)time

u vS t2

and v

2uS t

4.22 MOTION CURVESGraphical analysis is a convenient method of

studying the motion of a particle. It can be effectivelyapplied to analyse the motion situation of a particle.The variations of two quantities with respect to oneanother can be shown graphically.

For graphical representation, we require twocoordiate axes. The usual practice is to take theindependent variable along X-axis and the dependentvariable along Y-axis. In a context, with time as oneof the variables, it is usually taken along x-axis (sinceit is independent) and the other variable is along y-axis.



4.22(i) DISPLACEMENT TIME GRAPHS (S - T GRAPHS)Graphs are drawn with time along x-axis and displacement along yaxissignificance : i) The slope of the tangent at any point gives the instantaneous velocity. ii) The slope of the chord between two positions gives average velocity.

y

OX

t

S

S0

Context Shape of graph Comment on shape of graph

1) Particle at rest

i) A straight line parallel to x-axis. ii) S0 indicates initial displacement

2) Particle with Uniform velocity, . S = 0 at t = 0 The equation of motion is S = ut

y

OX

t

S

A straight line with positive slope

3) Particle with uniform acceleration, S = 0 at t = 0 The equation of motion is

S = ut + 12

a t2

y

OX

t

S

A parabola with increasing slope

4) Particle with uniform acceleration, S = S0 at t = 0 The equation of motion is

S = S0 + ut + 12

at2

y

OX

t

S

S0

i) A parabola with increasing slope ii) Intercept on y axis is initial displacement

5) Particle with uniform retardation S = 0 at t = 0 The equation of motion is

S=ut + 12

at2 where a is negative

y

OX

S

t

A parabola with decreasing slope

6) Particle projected vertically Upwards. The equation of motion is

S = ut - 12

gt2

y

OX

S

t

i) A parabola. ii) Its slope decreases from +K to zero and there after from zero to K where K is instantaneous slope.



4.22(ii) VELOCITY TIME GRAPHS (V T GRAPHS) :Graphs are drawn with time along x-axis and velocity along yaxissignificance :i) The slope of the tangent at any point gives the instantaneous acceleration.ii) The slope of the chord between two positions gives average acceleration.iii) The area of the figure bounded by the graph, the time axis and the ordinates drawn at the initial andfinal positions on the time axis gives the displacement in the bounded time interval.


1) Particle at rest

y

OX

v

t

As v = 0, the graph is a straight line along x-axis

2) Particle with Uniform velocity (u) The equation of motion is V = constant = u

y

OX

v

t

u

i) A straight line parallel to x-axis ii) slope = 0 , acceleration = 0 iii) y intercept = initial velocity = u

3) Particle with uniform acceleration, with velocity = 0 at t = 0 The equation of motion is V = at

y

OX

v

t

i) A straight line with positive slope ii) Slope = acceleration = constant

4) Particle with uniform acceleration, with velocity = u at t = 0 The equation of motion is V = u + at

y

OX

v

t

u

i) A straight line with positive slope ii) Intercept on y axis = u iii) Slope = acceleration = constant

5) Particle with uniform retardation, with velocity = u at t = 0 The equation of motion is V = u + at where a is negative

y

OX

v

t

u

i) A straight line with negative slope ii) Slope = retardation = constant

6) Particle projected vertically upwards The equation of motion is V = u gt

i) A straight line with negative slope. ii) The net area bounded by the curve is zero. iii) Net displacement in the time of flight = 0. iv) OA = time of ascent = u/g. AB = time of descent = u / g. OB = time of flight = 2u / g.

y

Av

t

XBO

u

u



4.22(iii) ACCELERATION - TIME GRAPHS (A T GRAPHS) :Graphs are drawn with time along x-axis and acceleration along yaxisSignificance i) The slope of the tangent at any point gives instantaneous jerk.

ii) The area under the graph with x axis (time axis) gives the change in velocity in thebounded time interval (v u)


1) The particle is with constant acceleration a0

a

y

xtO

0a

The graph is a straight line parallel to xaxis Y intercept = acceleration = 0a

2) Particle with non-uniform acceleration and with a = 0 at t = 0

a

tx

y

The graph is with variable slope

4.23 GRAPHICAL TREATMENT KINAMATICAL EQUATIONS OF MOTION OF A BODY WITH UNIFORM ACCELERATIONConsider a particle moving with initial velocity

u and uniform acceleration a. Suppose v is itsvelocity after t seconds. Let S be its displacementin the time interval t. The velocity time graph is astraight line with positive slope. The graph is givenby the line AB.v

yB

C A(u)

Velocity

v - u

O Dtime

x

t

1) To show that v = u + atThe slope of velocity time graph gives the

acceleration of the particle.

Here, = = = = BC v - uslope tan aAC t

v - u = at (or) v = u + at

2) To Show that 21S = ut + at2

The area under the velocity time graph withXaxis (time axis) gives the displacement in thebounded time interval. Here the area bounded bythe line AB with x-axis gives the displacement

S = Area of rectangle OACD + Area of triangleABC

1 S = OA OD + AC CB2

1 v2 u t t u

1 v

t2

uut t at a

21

2 S ut at

Note 4.14 :

General Method to show that 212

S ut at

When the particle is moving with uniformacceleration,

u + vAverage Velocity = 2



Displacement = (Average Velocity) (time)

u + v

S 2

t

u +u + at

S 2

t

V = u + at

21 S ut + 2 at

Note 4.15 : To show that 2 2v - u = 2aSv = u + at v u at ----- (1)

u + vS = t2

2Sv + u =t

----- (2)

Using (1), (2)

(v u) (v + u) = (at) 2St

= 2aS

2 2 v - u = 2aSNote 4.16:

Motion of a body

constantvelocity Acceleratedmotion

uniformlyaccelerated

non-uniformlyaccelerated

s = ut

v = u + at dsvdt

; dv dva vdt ds

u vs t2

s v a

differentiation

Integration

v2u2 = 2as ds vdt 21S ut at

2 dv a dt

n1S u a n2

ads vdv Note 4.17: For two bodies in motion for same time

interval we can use equation of motion in therelative form such as

r r rv u a t 21

2r r rS u t a t and 2 2 2r r r rv u a s

SOLVED PROBL EM S BASED ONKINEMATICAL EQUATIONS

Problem : 4.1The displacement x of a particle at the instant whenits velocity v is given by v= 3 16x . Find i tsacceleration and intial velocity

Sol. v = 3 16x or v2 = 3x + 16 or v2 - 16 = 3x Comparingwith v2 - u2 = 2aS , we get , u = 4 units,2a= 3 or a = 1.5

units

Problem : 4.2I f nS 2 0.4n find intial velocity and acceleration

Sol. nS 2 0.4n

n0.4 0.4S 2 0.4n2 2

n1S 2.2 0.4 n2

Compairing it with n1S u a n2

U = 2.2 units a = 0.4 units

Problem : 4.3The two ends of a train moving with uni formacceleration pass a certain point with velocity u andv. Find the velocity with which the middle point ofthe train passes the same point.

Sol.s/2

v1 v

s/2

u

12 2v u 2a s / 2 2 12v v 2as / 2

12 2 2 12v u v v 12 2 22v v v

2 21 u vv

2

Problem : 4.4The veloci ty of a body moving wi th uni formacceleration of 0.3m/s2 is changed to 3m/s in certaintime. I f the average velocity in the same time is 30m/s then find the distance travelled by it in that time.

Sol. v u = at3 0.3t t 10s S = Vaverage

x time

30 10 300 m



Problem : 4.5 A body starts from rest and moves with uniformacceleration of 5 ms-2 for 8 seconds. From that timethe acceleration ceases. Find the distance covered in12s starting from rest.

Sol. The velocity after 8 seconds v = 0 + 5 8 = 40 m/sDistance covered in 8 seconds

010 5 64 1602

s m After 8s the body moves with uniform velocity anddistance covered in 4s with uniform velocity

v=vt= 40 4 = 160 mThe distance covered in 12 s =160 +160 = 320 m.

Problem : 4.6A scooter can produce a maximum acceleration of5m s-2 . Its brakes can produce a maximum retardationof 10 ms-2 . The minimum time in which it can cover adistance of 1.5 km is ?

Sol. If v is the maximum velocity attained, thenv2 - O2 = 12 5 S . Also , O2 - v2 = 22 10 S

2 2

1 2,10 20v vS S

S = S1 + S2 2 2 231500

10 20 20v v v or

2 1500 20 100003

v or v = 100 m s-1

Problem : 4.7The speed of a train is reduced from 60 km/h, to15 km/h, whilst it travels a distance of 450 m. I f theretrardation is uniform , find how much further it willtravel before coming to rest ?

Sol. Here 5 5060 /18 3

u m s

5 2515 /18 6

v m s

Using v2 = u2 + 2as , we get2 650 25 2 9 450

3 6

or 2125 /

36 12a m s

If s1 is the further distance travelled before coming torest, then

21 25 25 36 12 30

2 6 6 2 125vs ma

Problem : 4.8A particle is at x = + 5 m at t = 0, x = 7 m at t = 6 sand x = + 2 m at t = 10s. Find the average velocity ofthe particle during the intervals(a) t = 0 to t = 6s (b) t = 6s to t = 10s,(c) t = 0 to t = 10 s.From the definition of average velocity

2 1

2 1

x x xvt t t

(a) The average velocity between the times t = 0 to t = 6s

1 1 2 25 , 0, 7 6x m t x m t s

Hence 12 1

12 1

7 5 26 0

x xv mst t

(b) The average velocity between the timest2 = 6s to t3 = 10 s is

13 22

3 2

2 7 9 2.2510 6 4

x xv mst t

(c) The average velocity between times t1= 0 to t3=10 s is13 1

33 1

2 5 0.310 0

x xv mst t

Problem : 4.9Velocity and acceleration of a particle at time t = 0 are

2 3 /u i j m s and 2 4 2 /a i j m s respectively.Find the velocity and displacement of particle att = 2s.

Sol. Here, acceleration a 4i 2j m/s2 is constant. So, we can apply

v u at and 21s ut at

2

substituting the proper values, we get

v 2i 3j (2) 4I 2J 10I 7J m/s

and 21 s 2 2i 3 j 2 4I 2J 12I 10J2

m

Therefore, velocity and displacement of particle at t =

2s are 10i 7j m/s and 12i 10j m respectively.. Problem : 4.10

A rifle bullet loses 1/20th of its velocity in passingthrough a plank. What will be the least number ofsuch planks required to just stop the bullet ?

Sol.2

219 220

v v ax 0

2 - v2 = 2anx

Dividing , 2

2 22

119 19120 20

vnv v



20 20 400 10.3

20 19 20 19 39

11 planks so, the bullet shall stop in 11 th plank.

Problem : 4.11A car starting from rest, accelerates at the rate of fthrough a distance S, then continues at constant speedfor time t and declerates at the rate 1v ft to come torest. I f the total distance travelled is 15 S then S =

Sol. A B Cf v v f/2

Dt

22t15 S

1t

AB= S = 211 f t2

-------- (1) 1v ft

BC = 1ft t

CD =

221ftu

2a 2 f / 2

nS 2 0.4m S + f t1 t +25 =15 S

1ft t 12 S -------- ( 2 )Dividing (1) by (2)

1tt6

21 tS f

2 36

2ftS72

Problem : 4.12

A body covers 100cm in first 2seconds and 128cm inthe next two seconds moving wi th constantacceleration. Find the velocity of the body at theend of 8sec?

Sol. 1100 2u a.42

-------(1)

228 = 6u+ 1 .36 22 a

(1) 3 2 gives 72 = 12aa = 6 cm/s2

1100 2u 6 42

2u = 112 u = 56cm/sV = u+ at = 56 6 8

V 8cm / s

Problem : 4.13A body starts with initial velocity u and moves withuni form acceleration.when the veloci ty hasincreased to 5u, the acceleration is reversed indirection, the magnitude remaining constant. Findits velocity when it returns to the starting point?

Sol. AB = S u a

BC

5uA|V

a

BA

For AB 25 u2 u2 = 2as

2 2V 25u 2 a s 2 2 2V 25u 24u

V2 = 49u2 V= 7uV is opposite to u V 7u

Problem : 4.14A train starting from rest travels the first part of itsjourney with constant acceleration a, second part withconstant velocity v and third part with constantretardation a , being brought to rest. The average

speed for the whole journely is 78v

. For what fractionof the total time, the train travels with constantvelocity ?

Sol.1

1

1 1

1 17 72 28 8 2

vt vt vtv t tort t t t t

or 6t = t1

Now 16 3

2 8 4t t

t t t

Problem : 4.15A particle traversed half of the distance with aveloicty of V0. The remaining parts of the distancewas covered with velocity V, for half of the time andwith V2 for other half of the time. Find the meanvelocity of the particle averaged and the whole timeof motion

Sol :0 1 2

S/ 2 S/ 2V V V

t2

t2

Average velocity for the second half distance =

1 2t tv v2 2t t2 2

= 1 2v v

2



Average velocity for the first half distance = 0v( it is constant)

Average velocity for total path

=

1 20 0 1

1 2 1 2 00

v t2v 2v v v2

v v v v 2vv2

Problem : 4.16A particle traversed along a straight line for firsthalf time with velocity V0 . For the remaining part,half of the distance is traversed with velocity V1 andother half distance with veloctity V2. Find the meanvelocity of the pariticle for the total journey.

Sol.

t2

t2

0v x2

x2

For the first half time average velocity = V0

For the second half time average velocity = 1 2

1 2

2v vv v

Average velocity for total journey = 1 2

01 2

2v vvv v2

0 1 2 1 2

1 2

v v v 2v vaverage velocity

2 v v

Problem : 4.17A car is moving with a velocity of 20 m/s. The driversees a stationary truck ahead at a distance of 100 m.After some reaction time t the brakes are appliedproducing a retardation of 4 m/s2. What is the maxi-mum reaction time to avoid collision ?

Sol. The car before coming to rest2 2 2v u as covers distance s

20 20 2 4s 400 508

s m

The car covers 50 mTo avoid the clash, the remaining distance 100 - 50= 50 m must be covered by the car with uniformvelocity 20 m/s during the reaction time t

50 5020 2.520

t st

The maximum reaction time t = 2.5 s

Problem : 4.18A driver can stop his car from the red signal at adistance of 20m when he is driving at 36 kmph and41.25m when he is driving at 54kmph. Find hisreaction time.

2us ut 2a

10020 10t 2.252a

22541.25 15t2a

3.75 = 7.5t

t 0.5s Problem : 4.19

A car starts from rest and moves with uniformacceleration 'a' . At the same instant from the samepoint a bike crosses with a uniform velocity 'u'.Whenand where will they meet ? what is the velocity of carwith respect to the bike at the time of meeting?

Sol. Sr = ur t + 21

2 ra t

0= ut 212

at

2uta

2

bike2u 2uS u.t u.a a

22

bikeuSa

Vcar =at = 2uVcar w.r. t. bike at the time of meeting = 2u u

cbV u

Problem : 4.20A bus starts moving with acceleration 2m/s2. A boy96m behind the bus simultaneously starts runningwith a constant velocity of 20m/s. After what time hewil l be able catch the bus?

Sol.

u 20m / s u 0 2a 2m / s

BoyBus

96m busS21

2r r rS u t a t

2196 20 22

t t 2 20 96 0t t

on solving t 8S and t 12SHe can catch the bus at two instants 8s and 12s. After12 seconds the bus will always be ahead of the boy



Problem : 4.21Two bodies start moving in the same straight line atthe same instant of time from the same origin. Thefirst body moves with a constant velocity of 40 m/s ,and the second starts from rest with a constantacceleration of 4 m/s2 . Find the time that elapsesbefore the second. catches the first body. Find alsothe greatest distance between then prior to it and thetime at which this occurs.

Sol. When the second body catches the first , the distancetravelled by each is the same.

2140 42

t t or t = 20 SNow , the distance s between the two bodies at any

time t is s = 212

ut at

For s to be maximum, 0 0ds or u atdt

or 40 10ut s

a u

Maximum Distance

= 2140 10 4 10 400 200 2002 m

Problem : 4.22Two ships A and B are 10 km a part on a line runningsouth to north . Ship A farthar north is streamingwest at 20 km/hr and ship B is streaming north at 20km/hr . What is their distance of closest approachand how long do they take to reach it ?

Sol. Ships A and B are moving with same speed 20 km/hrin the directions shown in figure . It is a twodimensinal , two body problem with zero accelera-tion

Let us find BAv

B

Bv

N

E

AB=10km

Av

BA B Av v v

here , 2 220 20 20 2 /BAv km hr

i.e BAv

is 20 2 km/hr at an angle of 450 from east

towards north . Thus , the given problem can be sim-plified as A is rest and B is moving with BAv

in the

direction shown inTherefore , the mininum distance between the two is

0min sin 45S AC AB = 5 2km

and the desired time is B

A

BAv

C

5 220 2BA

BCtv

5 2BC AC 1 154

hr minutes

Problem : 4.23Two trains one travelling at 54 kmph and the otherat 72 kmph are headed towards one another along a

straight track. When they are 12 km apart, bothdrivers simultaneously see the other train and applytheir brakes. If each train is decelerated at the rate of1 ms-2, will there be collision ?

Sol. Distance travelled by the first train before comingto rest

22

15 40072 / 2 1 200

2 18 2us m

a

225 122.52

m

Distance travelled by the second train beforecoming to rest

2

25 40072 / 2 1 200

18 2s m

Total distance travelled by the two trains beforecoming to rest = s1 + s2 =122.5 + 200 = 322.5 mBecause the initial distance of separation is 500 mwhich is greater than 322.5 m, there will be nocollision between the trains.

Problem : 4.24In a car race, car A takes time t less than car B andpasses the finishing point with a velocity v more thanthe velocity with which car B passes the point.Assuming that the cars start from rest and travel withconstant accelerations a1 and a2, show that 1 2

va a

t .

Sol. Let s be the distance covered by each car. Let the timestaken by the two cars to complete the journey be t1and t2, and their velocities at the finishing point be v1and v2 respectively.According to the given problem,

1 2 2 1v v v and t t t

Now,

1 2

2 1

v vvt t t =

1 2

2 1

2a s 2a s

2s 2sa a

=

1 2

2 1

a a

1 1a a

1 2V a at



*Problem : 4.25A particle moving along a straight line with initialvelocity u and acceleration a continues its motion forn seconds. What is the distance covered by it in the lastnth second ?

Hint. S = ut + 12 at

2

Displacement in n seconds = un + 12 an

2

Displacement in (n 1) seconds

= u ( n 1) + 12 a(n1)

2

Displacement in n th second = Displacement in nseconds displacement in (n 1) seconds.

12n

S u a n .

PROBLEMS BASED ON GRAPHS

Problem : 4.26A bus accelerates from rest at a constant rate forsome time, after which it decelerates at a constantrate to come to rest. I f the total time elapsed is tseconds then, evaluate.(a) the maximum velocity achieved and(b) the total distance travelled graphically.

Sol. (a) Let t1 be the time of acceleration and t2 that ofdeceleration of the bus.The total time is t = t1 + t2.Let vmax be the maximum velocity.As the acceleration and deceleration are constants thevelocity time graph is a straight line as shown in thefigure.with +ve slope for acceleration and -ve slopefor deceleration.From the graph,

the slope of the line OA gives the acceleration .

= slope of the line OA = max max11

v vtt

the slope of AB gives the deceleration

= slope of AB = max max22

v vtt

t = t1 + t2 = max maxv v

tt1 t2O

A

B

v

vmaxmaxt v

maxv tab

a b

(b) Displacement = area under the v-t graph

= area of OAB

max1 12 2

base height t v

= 12

tt

= 21

2t

Problem : 4.27

Figure shows the motion of a particle along a straightline. Find the average velocity of the particle duringthe intervals(a) A to E; (b) B to E; (c) C to E;(d) D to E; (e) C to D.

Sol.

12

4

8

5 10 tA

B

CD

E

O 83

x cm

(seconds)

(a) As the particle moves from A to E, A is the initialpoint and E is the final point.The slope of the line drawn from A to E

i.e., tx

gives the average velocity during that interval

of time.

The displacement x isxE xA = 10 cm 0 cm = +10 cm

The time interval EAt = tE - tA = 10s.

During this interval average velocity

110v 110

x cm cmst s

(b) During the interval B to E, the displacement

cmcmcmxxx BE 6410 and

.7310 sssttt BE

Average velocity 6v7

x cmt s

= +0.857 cms1 = 0.86cms1



(c) During the interval C to E,the displacement10 12 2E Cx x x cm cm cm and

1

10 5 52 0.45

E Ct t t s s sx cmv cmst s

(d) During the interval D to E, the displacement

cmcmcmxxx DE 21210 and the time interval

10 8 2E Dt t t s s s 11

22

cmsscm

txv

(e) During the interval C to D,the displacement01212 cmcmxxx CD

and the time interval sssttt CD 358

The average velocity 10

30

mssm

txv

(The particle has reached the same position duringthese 3s. The average velocity is zero because thedisplacement is zero).

*Problem : 4.28

Velocitytime graph for the motion of a certain bodyis shown in Fig. Explain the nature of this motion.Find the initial velocity and acceleration and writethe equation for the variation of displacement withtime. What happens to the moving body at point B ?How does the body move after this moment ?

Sol.

A

B ts

V m/s

C0

5

5 10 15

The velocity time graph is a straight line withveslope. The motion is uniformly retarding upto point Band there after uniformly accelerated upto C.At point B the body stops and then its direction ofvelocity reversed.The initial velocity at point A is v0 = 7 ms

-1.

1

20 7 711 11

f ov v msa mst s

= 0.64ms

2

The equation of motion for this body which gives

variation of displacement with time is 217 0.642

S t t

=7t - 0.32t2.

*Problem : 4.29The graphs in (i) and (ii) show the S t graph andV t graph of a body.Are the motions shown in the graphs represented byOAB the same ? explain

(i)

O 5 10B

x cm

5

10A

f, seconds

(ii) 5

O 5 10

v cm/s

t, seconds

A

Bi

Sol. The motion shown by the two graphs are not same.i) In the given s t graph OA, is a uniform retardationmotion.

Here,

displacement = (average velocity) x (time )

10 02u 4

u = 5ms1using v2 u2 = 2as

0 52 = 2a(10)

a = 1.25 ms 2

ii) In the given V t graph, OA is a uniform retardationof motion

a = slope of the line 1

24 14

OA ms msOB s

Thus the two graphs even though represent uniformretardation motions, the magnitudes are not equal.



*Problem : 4.30The acceleration - displacement graph of a particlemoving in a straight line is given as in the fig. Theinitial velocity of the particle is zero. Find the velocityof the particle when displacement of the particle iss = 12m. a(ms2)

2

4

2 8 10 12 S(m)

Sol. From the equation v2 - u2 = 2as

as = 2 2

2v u = area under a-s graph

initial velocity u = 0;2

2vas = area under a-s graph.

2 V area under a s graph

11 1 12 2 6 2 2 4 2 2 4 2 2 24 4 32 2 2

ms

*Problem : 4.31A body starts from rest and travels a distance S withuniform acceleration, then moves uni formly adistance 2S and finally comes to rest after movingfurther 5S under uniform retardation. Find the ratioof average velocity to maximum velocity :

Sol. Gr aphi cal l y : Area of (Vt) curve representdisplacement

V

t1 t2 t3

Vmax Vmax

t

S = 12

Vmaxt1 or t1 = max2S

V

2S = Vmax t2 or t2 = max2S

V

5S = 12

Vmaxt3 or t3 = max10SV

avTotal displacement

VTotal time

av

max max max

S 2S 5SV

2S 2S 10SV V V

av

max

V 8S 4V 14S 7

(OR)

2max

Total displacementavtotal displacement Displacementduringacceleration Duringuniformandretardaton velocity

VV

av

max

V 8S 8 4V 2 S 5S 2S 14 7

Problem : 4.32Figure given here shows the displacement time graphfor a particle. Is it practically possible ? Explain.

Dis p

lac e

me n

t

time

Sol. From the graph, it is evident that, at any instant oftime the particle possesses two displacements, whichis impossible.

Problem : 4.33Figure given here shows the variation of velocity of aparticle with time.

0 2 4 7t2

8

v ms-1

4

Find the following :i) Displacement during the time intervals.a) 0 to 2 sec. b) 2 to 4 sec.and c) 4 to 7 secii) Accelerations ata) t = 1 sec, b) t = 3 sec. andc) t = 6 sec.iii) Average accelerationa) between t = 0 to t = 4 sec.b) between t = 0 to t = 7 sec.iv) Average velocity during the motion.Hint.i)displacement = Area enclosed between v t graphand time axis.ii) Acceleration = slope of v t curve



iii) Total change in velocityAverage acceleration =Total time

iv) Average velocity = timeTotalntdisplacemeTotal

Ans. (i) a) 8m b) 16 m c) 12m

ii) a) 4ma2

b ) 0 c ) 2/3ms2

iii) a) 2m/s2 b) 0

iv) 1367 ms

Problem : 4.34

The velocity time graph of a moving object is givenin the figure. Find the maximum acceleration of thebody and distance travelled by the body in the intervalof time during which this acceleration exists.

80604020

Vel

o city

(in

m/s )

0 10 20 30 40 50 60 70 80Time (in)

Sol . Acceleration is maximum when slope is

maximum 2max80 20a 6m / s40 30

S = 20 m/s 10s +12

x 6m/s2 100 s

2 = 500 m

Problem : 4.35

The velocitytime graph of a body moving in astraight line is shown in Fig. Find the displacementand distance travelled by the body in 10 sec.

t(s)

20

100

1020

u(m/s)A

D

C

2 4 6 8 10

Hint. The area enclosed by velocity-time graph withtime axis measures the displacement travelled in thegiven time.Ans. S = 50m

-2

Problem : 4.36S t group of a particle moving an a straight line isas shown. On which part the force acting is zero.

Sol : dsvdt

constant in the part bc a 0

S

b

c

d

tl

0

F 0 in the part bc

Note 18 : The acceleration, time graph is as shown

t1 t2t

a

The corresponding vt graph will be

t1 t2t

v

Problem : 4.37

Three particles start from the origin at the same time,one with a velocity v1 along x - axis , the secondalong the y - axis with a velocity v2 and the thirdalong x = y line. The velocity of the third so that thethree may always lie on the same line is

Sol. Let time interval be chosen as 1 second.

0x

y

vPA OAPB B v

So , P ( x ,y) divides AB in the ratio vx : vy0x y x x yx y x y

v v v v vx

v v u v

p (x,y)

0A

B(0,vy)

vy

vx

y= 0x y y

x y

v v vv v

x y

x y

v vv v



2 2v x y

2 x yx y

v vv v

Now, repalce vx by v1 and vy by v2

Problem : 4.38The displacement - time graphs of two particles Pand Q are as shown in the figure. The ratio of theirvelocities Vp and VQ will be

600

300

Sol. The velocity of a particle is equal to the slope of time- dispalcement straight line.

0

0

1tan 30 3 1: 3tan 60 3

P

Q

VV

Problem : 4.39The a - t graph is shown in the figure. The maximumvelocity attained by the body will be

11

10

0

a (m

s-2 )

t(s)

Sol : Maximum velocity = at=Area between v-t graph and t-axis

1 11 10 552

m

Problem : 4.40A car travels starting form rest wi th constantacceleration a and reaches a maximum velocity V..I t travels with maximum velocity for some time andretards uniformly at the rate of b and comes to rest.I f s is the total distance and t is the total time ofjourney then t =

Sol :1t 2t 3ta b

1S 2S 3SV V

1vV 0 t,; taa

2

21 1

1 vS t2 2

aa

3 3vO V t ; tbb

2 2

3O V 2 Sb

223 3

v 1S t2 2

bb

v

t1t 2t 3t 1 322

S S SStV V

2v 1 1S2

Va b

1 2 3t t t t

= v s v v v

v 2 2a a b b

s vtv 2

a bab

2 1 2s s s s

2 2v vs

2 2a a

2vs

a

s vt s 2s 2s vt

[S2= distance travelled with constant velocity]

PROBLEMS BASED ON CALCULUS Problem : 4.41

A point moves the plane x - y according to the lawx = k sin t and y = k(1cos t ) where k and arepositive constants. Find the distance s traversed by theparticle during time t.

Sol. xdx

v k cos tdt

and ydy

v k sin tdt

Now, speed v = 2 2x yv v k constant s = vt = k t .

Problem : 4.42The coordinates of a body moving in a plane at anyinstant of time t are 2x ta and 2y tb .The velocity of the body is



Sol. 2x ta 2xdxv tdt

a

2y tb . 2ydyv tdt

b

2 22 2 2 2x yvelocity v v v t ta b 2 22t a b

Problem : 4.43

The motion of a particle along a straight line is de-scribed by the function 2 46 4s t t in SI units.Find the velocity, acceleration, at t=2s, and the aver-age velocity during 3rd second.

Sol. 2 46 4s t t

Velocity 38 4 2ds t t when tdt

Velocity = 8 2 -4 23

Velocity = 16 m/s

Acceleration 2

22 8 12

d sa tdt

when t=2

acc = 8-12 22 = -40

acc = -40 m/s2

displacement in 2 secondss1 = 6 + 4. 2

2 -24 = 6 mdisplacement in 3 secondss2 = 6+ 4. 3

2 - 34 = -39 mdisplacement during 3rd second= s2

- s1 = -39 - 6 = -45 m Average velocity during 3rd second

45 45 /1

m s

-ve sign indicates that the body is moving in oppositedirection to the initial direction of motion.

Problem : 4.44

A par ticle moves according to the equation3 t x , where will be the particle come to the

rest for the first timeSol. x = (t 3) 2

x= t2 6t + 9

2 6dxv tdt

0 = 2t 6

t 3s

Problem : 4.45Acceleration of a particle is varying according tothe law a = ky. Find the velocity as a function of y,and intial velocity V0

Sol.dv dv dyadt dy dt

0

yV2 20

V 0

v dv kydy V V ky 2 20V V ky

Problem : 4.46The velocity of a particle moving in the positivedirection of the X-axis varies as V = K S where K is apositive constant. Draw V t graph.

Sol. V = K s

dS K Sdt

S t

0 0

dSK dt

S

2 S = Kt and S = 2 21K t4

2 21 1K 2t = K4 2

dSV tdt

V t The V t graph is a striaght line passing throughthe origin

Problem : 4.47In the arrangement shown in figure the ends of aninextinsible string move downwards with uniformspeed u. Pulleys A and B are fixed. Find the speedwith which the mass M moves upwards.

O

Mo o

A B

Sol. Suppose the distance of point O from the ceiling is y andthe distance of point O from each pulley is x and thedistance between the two pulleys is .

22 2

4lx y x

/ 2l

y

Differentiating the above eqation w.r.t to t

12 2 24

dx dy dx ydt dt dt



But, dx

udt

, v = dydt

and ddt

xu = yv and v = uxy = u sec

Velocity of mass = v = u sec (upwards) Problem : 4.48

Figure shows a rod of length resting on a wall and thefloor. Its lower end A is pulled towards left with a constantvelocity v. Find the velocity of the other end B downwardwhen the rod makes an angle with the horizontal.

A

B

V

Sol. In such type of problems, when velocity of one part ofa body is given and that of other is required, we firstfind the relation between the two displacements, thendifferentiate them with respect to time. Here if thedistance from the corner to the point A is x and up to Bis y. Then

v= dxdt

& vB=dydt

(sign denotes that yis decreasing)

Further, x2 + y2 = 2Differentiating with respect to time t

2x dx dy

2y 0dt dt

xv = yvB

vB = (v) xy =v cot

Problem : 4.49

Accelerati on of a par ti cl e at any time t i s 2 2 2 3 /a ti t j m s . I f initially particle is at rest,

find the velocity of the particle at time t = 2s.Sol. Here acceleration is a function of time, i.e.,

acceleration is not constant. So, we cannot applyv u at

.

We will have to go for integration for finding velocityat any time t.

dva =dt

Thus dv a

dt

or v 2

0 0

dv adt

or 2

2

0

v 2ti 3t j dt

= 22 30

t i t j = 4i 8j m/sTherefore, velocity of particle at time t = 2s is

4i 8 j m/ s

Problem : 4.50A point moves the plane x y according to the lawx = k sin t and y = k (1cos t ) where k and arepositive constants. Find the distance s traversed by theparticle during time t.

Sol. xdx

v k cos tdt

and ydy

v k sin tdt

Now, speed v = 2 2x yv v k constant s = vt = k t .

Problem : 4.51t = ax2 + bx find acceleration ?(a, b are constants)

Sol. t = ax2 + bxdt 2ax bdx

dx 1vdt 2a b

2dv 2a dxdt dt2ax b

32a 2av

2ax b

Problem : 4.52

A point moves rectilinearly with deceleration whosemodulus depends on the velocity v of the particle as

,k va where k is a positive constant. At the initialmoment the velocity of the point is equal to V0. Whatdistance will it take to cover that distance?

Sol. Let t0 be the time in which it comes to a stop. Given

that dv k Vdt

0

0 0

0t

v v

dvkdtv

0 02kt v

0 02t vk

Now to find the distance covered before stopping,

dv dv ds dvvdt ds dt ds



But, ;dv k Vdt

dvv k Vds

vdv kds

0

0 32

00

23

s

V V

vdv k ds s Vk

Problem : 4.53A par ticle moves according to the equationdv vdt

a b where a and b are costants. Findind

velocity as a function of time. Assume body startsfrom rest.

Sol.dv vdt

a b v t

0 0

dv dtva b

0

Vln v

ta b

b ln v ln ta b a b vln ta b ba

tv e ba ba

1 tv e bba

1 tv e bab

Problem : 4.54The accelerati on a of a parti cle depends ondisplacement S as a = S + 5. I t is given that initiallyS = 0 and V = 5 m/s.Find relation between i) V and S ii) S and t

Sol. i) : a = dV dV dS. S 5dt dS dt

dVV 5ds

S

V S

5 0

V dV S 5 dS

2 2

5 0

52 2

V SV S S

V = (S+5)

ii) V = S + 5

ds 5dt S

0 0

5

S tdS dtS

se 0

log S 5 t , loge S 5

5

= t

Problem : 4.55

The x and y co-ordinates of a par ticl e are sinx A tw and sin / 2y A tw p . Find the

motion of the particleSol. Given sinx A tw

sin / 2 cosy A t A tw p w squaring and adding 1) and 2) we get

x2+y2=A2

Y

X3 / 2t / 2t

t

0t

i.e . path of the particle is a circle with centre atorigin and radius AAt time 0tw x = 0 and y=Aand at / 2t x Aw p and y=0at 0t xw p y = A and so on. The motion is circular clockwise

Problem : 4.56An object is projected in X Y plane in which velocitychanges according to relation V ai bxj

. Equation

of path of particle is:a) Hyperbolic b) Circularc) Elliptical d) Parabolic

Sol.dx

adt

x = atVy = bx = b at

dyb

dt at

dy bat dt y =

2bat2

y = 2 2

2ba x bx2 2aa

y 2x i.e., parabolic.



Problem : 4.57The displacement (x) of a particle varies with timeas x = ae t + be t where a, b, , are positive constant.Find how does the velocity of particle change with time,

Sol : x = ae t +be t

t tdxV ae bedt

V = t tbe ae V increases as t increases.

4.24 ACCELERATI ON DUE TO GRAVI TY (G)The uniform accleration of a freely falling

body towards the centre of ear th due to ear thsgravitational force is called acceleration due togravity.i) It is denoted by gii) Its value is constant for all bodies at a givenplace. It is independent of size, shape, material,constitution(hollow or solid), nature of the body. Ifair resistance is ignored, all the bodies as light as afeather to a heavy metal sphere, droppedsimultaneously from the same height hit the floor atthe same time because all the bodies have sameacceleration due to gravity.iii) Its value changes from place to place on thesurface of the earth.iv) It has maximum (greatest) value at the poles ofthe earth. The value is nearly 9.83 m/s2.

It has minimum (least ) vlaue at the equator ofthe earth. The value is nearly 9.78 m/s2.v) The average value of g on earths surface is 9.8 m/s2.

vi) On the surface of moon, 21.67 / 6earthgg m s

On the surface of sun, g = 274 m/s2

vii) The acceleration due to gravity of a body isalways directed downwards towards the centre ofthe earth, whether a body is projected upwards ordownwards.viii) When a body is falling towards the earth, itsvelocity increases, g is positve.ix) When a body is projected upwards, its velocitydecreases, g is negative.x) The acceleration due to gravity at the centre ofearth is zero.

4.25 EQUATIONS OF MOTION FOR FREELYFALLING BODYMotion of all the bodies falling towards the Earth

when air resistance is ignored is known as free fall.For a freely falling body, u = 0, a = +g (i) v u at v = gt

(ii) 212

S ut at 21

2S gt

(iii) 2 2v 2u aS 2v 2gS

(iv) 12n

S u a n

12nS g n

4.26 EQUATIONS OF MOTION FORVERTICALLY PROJECTED BODYFor a body projected vertically upwards,a = g (since velocity, acceleration vectors are

opposite) (i) v vu at u gt

(ii) 2 21 12 2

S ut at S ut gt

(iii) 2 2 2 2v - u = 2aS v - u = 2 - g S

(iv) 1 12 2n n

S u a n S u g n

4.27 MOTION PARAMETERS OF A BODYPROJECTED VERTICALLY UPWARDSi) Maximum height (H

max) :

For a body projected vertically upwards, themaximum vertical displacement from ground aboutwhich its velocity is zero is called its maximumheight.

Expression :Let a body be projected vertically upwards with

initial velocity u.We know that, v2 u2 = 2ashere a = g, s = Hmax, v = 00 u2 = 2 ( g) Hmax u2 = 2gHmax

2

max 2uHg



(ii) Time of ascent (ta) :

For a body projected upwards the time to reachthe maximum height is called time of ascent

Expression :Let a body be projected vertically upwards with

initial velocity u.We know that, v = u + atHere a = g, t = ta, v = 0

0 au gt au gt

autg

(iii) Time of descent (td) :For a body projected upwards the time to travel

from maximum height to the point of projection onground is called time of descent

Expression : Let a body be projected verti-cally upwards with velocity u.

Step 1) : For upward motion

2 max2

max

0 2 H

2

u g

uHg

0initialvelocity

maxS H

u

Step 2) : For downward motionS = (initial velocity) t + 2

12

gt

Here,

initial velocity = 0, ,a g ,dt t maxS H2

max1 02 d

H gt 2

2

22

2

1 2 2

d

d

u gtg

utg

dutg

Note 4.19 :For a body projected vertically upwards,Time of ascent (ta) = Time of descent (td)=

ug

iv) Time of flight ( t f ) :For a body projected vertically upwards the

sum of time of ascent and time of descent is calledtime of flight (tf) It is the total time for which thebody remains in airTime of flight = Time of ascent + Time of descent

f a dt = t + t

f2t u u u

g g g

f2 t ug

v) Velocity of the body on reaching the pointof projection (V

str iking)Let a body be projected vertically upwards with

initial velocity u.The body reaches the point of projection once

again after the time of flight (tf)We know that, V = u + atHere,

a = g, t = tf = 2ug , V = VVstriking

striking2u V = u - gg

u

striking V = u

The body reaches the point of projection withthe same speed of projection but in opposite direction.Note 4.19 : For a body projected vertically upwards s-t , v-t , a-t graphs are as follows :

2

2ug

s

ug

2ug

t

v

t -g

ta

Note 4.20 : For a freely falling body s-t , v-t , a-t graphs are as follows :

s

t

v

tq

tan gq

a

t



Note 4.21: In the case of a vertically projected body,velocity is maximum at projection point and mini-mum (zero) at highest point

Note 4.22 : Velocity goes on decreasing .

Note 4.23: Velocity at any point during the upwardjourney = velocity at the same point during thedownward journey (numerically)

Thus projection velocity = Striking velocity (nu-merically) .

Note 4.24: Change in velocity in the entire journey= 2u {( 2f iv v v u u u }Note 4.25: Similarly Change in momentum. duringthe entire journey = 2mu

Note : 4.26Velocity at half maximum height = 2

u

Sol. using the equation v2-u2=2as, we have a=-g ,21 1

2 2 2us Hg

, we get

2 2 2

2 2 2 2 224 2 2 2u u u uv u g v u v v

g

Note 4.27:

Velocity at 34

th of maximum height = 2u

Sol. using the equation v2-u2=2as, we have a=-g ,23 3

4 4 2us Hg

, we get

2 2 2

2 2 2 2 23 328 4 4 2u u u uv u g v u v vg

Note 4.28 : Distance covered by a body projectedvertically up in the 1st second of its upward journey= Distance covered by it in the last second of itsentire journey = 2

guSol. we know that for a body projected vertically

up, 12n

s u g n .

Substituting n=1 gives 1 2gs u (clearly , the

distance travelled in last second is same as that of 1stsecond )

Note 4.29:Distance covered by a body projected vertically

up in the last one second of its upward journey =Distance covered by it in the 1st second of its downward

journey = 2g

Sol. For a body falling downwards , we know that21

2s gt

Substituting t=1 gives us 2gs (clearly , the

distance travelled in1st second in the downwardjourney is same as that of last second of upwardjourney )Note 4.30:

Time taken by vertically projected up body to

reach 34

th of maximum height = 2at

Sol. we know that , for a body projected vertically

up v=u-gt 2u u gt (since at 34 th of maximum

height , velocity = 2u

)

12 2 2 au ugt t t t

g

Note 4.31:A body projected vertically up takes a time

112a

t t to reach half of maximum height .

Solution : we know that , for a body projectedvertically up v=u-gt . 2

u u gt (since at half ofmaximum height , velocity = 2

u)

1 11 12 2 2a

u uu gt t t tg

Note : 4.32When air resistance is taken into accounti) Time of ascent is less than that in vacuumii) Time of ascent is less than time of descentiii) The speed of the body when it reaches thepoint of projection is less than the speed ofprojection



R

O

u

mg

u

v

R

mg

F = mg + R F = mg RRa gm

1 Ra g m 0 = u ata v= 0 + a

1td

aut Rg m

d

Rv g tm

2 2a d

1 R 1 Rh g t g t2 m 2 m

a

d

Rgt vmRt ug m

d

d

R Rg gtv m mR Ru tg gm m

For dropped bodiesi) Same resistance force R

/a g R m If m is more a is more heavier body falls firstii) If R is proportional to m then acceleration issame for both

both the balls fall simulateneously in thesame timeiii) If m is same R is less for smaller body; a =gR/m is more for smaller body smaller body falls first

Application : 4.11Body 1 is released from the topof a tower . At the same instant, h

u=0

2

4

1

body 2 is projected ver tically upas shown then

a) height at which they meet is ht = uSol. Let the two meet after a time 't' seconds then

the in this the distance covered by both mustbe equal to height of true

i.e , S1+S2=h

2 21 12 2

gt ut gt h ut h htu

b) the time after which their velocities areequal is 2

utg

Sol : Let the velocities be equal after a time 't'

1 2v v

2ugt u gt u tg

c) Ratio of distances covered when themagnitudes of their velocities are equal is

1 2: 1:3S S Sol : From above , velocities are equal after a time

2utg

in this time2 2

1 2

1 12 2 2 4

u uS g gg g

2

1 8uSg

2

22

1 12 2 2 2

u uS ut gt u gg g

= 2 2 23

2 8 8u u ug g g 1 2S :S = 1: 3

Application : 4.12Three bodies are projected from towers of

same height as shown. 1st one is proj ectedver tically up with a velocity 'u' . The second oneis thrown down ver tically with the same velocityand the third one is dropped as a freely fallingbody. I f t1, t2, t3 are the times taken by them toreach ground, then,



a) velocity of projection is 1 212u g t t

h h h

1 2

u=u

3

u=0

Sol. Clearly the extra time taken by the 2nd bodyis equal to the time of flight of 1st body

i.e, 1 2 1 22 12ut t u g t tg

b) height of the tower is 1 212

h gt tSol. We know that , for a vertically projected upbody

212

s ut gt 21

2h ut gt [similar to ax2+bx+c=0]

The product of these far roots of theirchapton gives

1 21h = gt t2

c) The time taken free fall in the 3rd case

is given by 1 2t t t 2htg

in above , the form that

1 2 1 21 22

uh gt t t tg

1 22

freeht t t

g

Application : 4.13A body falls freely from a height 'H'. After

t seconds of fall, gravity casses to act. Find thetime of flight

Sol. Let the total time taken be T

Let gravity cease at P

Q

O U = 0

P

x

H

i,e . at , 0 tanP g v cons t Step a] Distance converd in t second is

212

gt x

remaining distance PQ is to be convered bythe body with constant velocity for which haveSPQ = VP x t

'

Step b] Velocity at P is VP = gtStep c] SPQ = VP x t' (t'=time taken to cover the

distance PQ)

' PQP

S H xt

V gt21

22

H gt H tt

gt gtiv) total time of fall is

'

2H tT t t tgt

Application : 4.14

A par ticle is projected ver tically upwardsand it reaches the maximum height H in time Tseconds. The height of the par ticle at any time twill be.

212H g t T

Sol. From v=u+gto = u-gtu = gT ..............i

221 .......

2 2uH gT iig

Let h be the distance travelled in time t.

Then 212

h ut gt 21 ........

2h gTt gt iii Subtracting (ii) from (iii)

2 21 12 2

h H gTt gt gT

22 222 2g gTt t T T t

212

h H g T t



Application : 4.15Roket i s f i r ed ver t i cal l y up with an

acceleration a. Fuel is exhausted after t sec.M aximum height i t can r each is given by

21 12

aa tg

Sol. Step I :- Distance covered by the rocket in tsec. (till fuel is exhausted)

2112

S at

Step I I : Velocity at the end of t sec,v=u+at = 0+at = at (u = 0 for a

rocket)Step III:- Further distance it can go up after t

sec.(after fuel is exhausted) is given by

stopping distance 2

2 2velocity

Sretardation

=

222 2

atvg g =

2 2

2a t

g

Step -IV = Maximum height the rocket carreach is , H= S1 +S2

2 22 21 1 1 1

2 2 2a t aH at at

g g

Note : 4.33For a Vertically thrown up body, maximum

height H = 218

gT where 'T' is the time of flight

Sol. We know that time of flight 2uTg

2gTu

Maximum height2

2212

2 2 8

gTuH gTg g

Similarly the same formula is applicable evenin the case of a projectile.

Application : 4.16After falling for t1 sec, a stone hits a hor i-

zontal glass plate, where it looses 50% of its ve-locity. I t then takes t2 sec to reach ground. Findthe Height of the glass plate above the ground.Sol. Let P be the upper edge and Q be the loweredge of the glass plate, as shown.

Step -I :- Velocity of the stone at P, Vp = gt1

Step -II:- Velocity of the stone at Q, vQ = 1

2gt

( it looses 50% of velocity)Step - III: from Q to R, the stone travels as a

vertically thrown down body with an initial velocity

= 12gt

u = 0 O

t1 sec

P

Q

t2 sce

R

Step -IV : Hence using the equation,

s=ut+ 212

at

Where S = QR ; 1 ;2Qgtu v a g

t = t2, we have,

height of the glass plate above the ground =21

2 21

2 2gtQR t gt . From this equation, t2

can be found.Application : 4.17

A fr ictionless wire is fixed between A andB inside a sphere of radius R . A small ball slipsalong the wire.Find the time taken by the ball toslip from A to B

A

BR

Oq

900

O

R

qg cos

A

B

R

g

Sol. 21

2S at



i.e. AB 21 cos 2 cos2 g t Rq q

2 Rtg

4.28 VERTICAL PROJECTION OF ANOBJECT FROM A TOWER(EXPRESSION FOR HEIGHT OF TOWER)Consider a tower of height H. Suppose a body

is projected upwards vertically with initial velocityu from the top of tower. Suppose it reaches to a dis-placement x above the tower and there after reachesthe foot of the tower. Let t be the total time of travel.

Now, considering the total path of the body,the motion parameters are as follows.

Initial velocity of the body = uNet displacement of body = S = +xxH= HTime of travel = t

we know that, 212

S u t a t Here,

2

2

a = - g, S = - H, u = u, t = t 1 gt2

1 gt2

H ut

H ut

TOWER

H

x

H

x

u

Note : 4.34This is a quadratic equation in time.Comparing with the standard quadraticequation

2ax + bx + c = 0

we get 2u u + 2gH

t = g

Application : 4.18 A par ticle projected ver tically up from the

top of a tower takes t1 seconds to reach theground. Another par ticle thrown downwardswith the same velocity from the same point takest2 seconds to reach the ground. Then

a) Velocity of projection = 1 22gu t t

b) Height of tower = 1 212

H g t t

c) A body dropped from the top tower takestimes 1 2t t t to reach the ground

1 212

H g t t 212

g t

1 2 t t tHint.For the body projected vertically upwards,

21 1

12

H gt ut ------- (1)

For the body projected vertically downwards

22 2

12

H ut gt -------- (2)

Solve (1), (2)

Application : 4.19A body projected ver tically upwards from

ground is at the same height h from the groundat two instants of time t1 and t2 (both beingmeasured from the instant of projection) Now

a) 1 21h = g t t2

b) Velocity of projection = 1 21u = g t + t2

c) H max 2

1 21= g t + t8

d) A body dropped from height h takes time1 2t t to reach the ground

Problem : 4.58A body is projected vertically up with velocity u from atower. I t reaches the ground with velocity nu. The height

of the tower is 2

2 12uH ng

Sol. v2u2=2as Here u = u, v = nu, a = g, s = H

2 2

2 2

2

1 2

nu u gH

n u gH

22 1

2uH ng

Problem : 4.59Two bodies begin to fall freely from the same height.The second body begins to fall t s after the first. Afterwhat time from the begining of first body dose thedistance between the bodies equals to ?



Sol. Let the time of fall of the 1st body be t seconds . Time offall of second body = t .Distances of free fall of the bodies in the above timeintervals respectively are

2

;2

2

2

2

1

tgHgtH

Therefore 221 21

ggtHH

2

gt

Problem : 4.60

One body falls freely from a point A at a height (H + h)while another body is projected upwards with an initialvelocity V0 from point C at the same time as the firstbody begins to fall. What should be the velocity V0 ofthe second body so that the bodies meet at a point B at theheight h? What is the maximum height attained by 2ndbody for the given initial velocity? What is the value ofV0 if H = h ?

Sol.

H+h

B

Ch

H

A

i) Suppose the two bodies meet after t seconds.2

2gtH .....(1)

2

2ogth V t ........ (2)

Solving (1) & (2)

2ogV H hH

ii) We know that 2

2MaxuHg

2 22max max( )2 2 2 4

o H h H hV gH Here H hg g H H

iii) When H = h, we get Vo= 2gh .

Problem : 4.61For a freely falling body, Find the ratio of the timestaken to fall successive equal distances.

Ans. Ratio of times taken to fall equal distances is

1 0 : 2 1 : 3 2 :........ : 1n n

Hint. use t = 2hg

Problem : 4.62I f a freely falling body covers half of its total distance inthe last second of its journey, Find its time of fall.

Sol. Suppose t is the time of free fall.21

2h gt ....... (1)

21 12 2h g t ............ (2)Solving 1, 2

2 2t s since 2 2 is not acceptable.

Problem : 4.63

A balloon starts from rest, moves vertically upwardswith an acceleration g/8 ms-2 . A stone falls from theballon after 8s from the start. Find the time taken by thestone to reach the ground (g = 9.8ms2)

Sol.Step1 : To find the distance of the stone above theground about which it begins to fall from the balloon.

212

S ut at

here, s = h , u = 0 , a = g/8

21 8 42 8

gh g Step2 : The velocity of the balloon at this height can beobtained from v = u + at

0 88gV g

This becomes the intital velocity (u|) of the stone as thestone falls from the balloon at the height h.

u| = gStep-3 : For the total motion of the stone 2 |

12

h gt u t Here, h = 4g , u| = g, t = time of travel of stone.

214 2g gt gt 2 2 8 0t t

solving for t we get t=4 and 2s. Ignoring negativevalue of time, t=4 s



Problem : 4.64A rocket is f i red upwards vertical l y wi th a netacceleration of 4m/s2 and initial velocity zero. After 5seconds its fuel is finished and it decelerates with g. Atthe highest point its velocity becomes zero. There afterit accelerates downwards with acceleration g and returnback to ground.i) Plot velocity time graph for complete journeyii) Displacementtime graph for the complete journey.( Take g = 10 m/s2.)

Sol. Stage i : To find velocity of rocket after 5 seconds 10 4 5 20A OAV at ms

Stage ii : To find further time of ascent after 5 seconds.0 = 20 g tAB

20 2seconds10AB

t

Here, the total vertical displacement of stage i) and stageii) is

= area of OAB = 1 7 202

= 70m.

A

B t(s)O 5 7 10.7

v(m/s)

20

C

s(m)

70

50

5 7t(s)

O 10.7

A

B

C

Stage iii : If tBC is time of descent then

70 = 12

(10) 2BCt tBC = 14 = 3.7s

Note 4.36 : OABCt 7 + 3.7 = 10.7sNote 4.37 : SOA = area under v t graph

= 12

(5)(20) = 50m

Problem : 4.65

A stone is allowed to fall from the top of a tower 300 mheigh and at the same time another stone is projectedvertically up from the ground with a velocity 100 ms1.Find when and where the two stones meet ?

Sol. Suppose the two stones meet at a height x from groundafter t seconds.

211002

x t gt .....(1) 21300 02

x gt .......(2)

Solve 1, 2t = 3 sec, x = 255.9m

Problem : 4.66Ball A is dropped from the top of a building and at thesame instant that a ball B is thrown vertically upwardfrom the ground. When the balls collide, they are movingin opposite directions and the speed of A is twice thespeed of B. At what fraction of the height of the buildingdid the collision occur ?

Sol. Given A BV = 2VLet 1h and 2h are the distances travelled by the two balls

22

1 222 ghugh 2

1 22 4 uh hg

......... (1)

If they meet after t seconds, for the condition VA = 2V

B

0 + gt = 2 (u gt)

gtugtugt 3222 gut 3/2

Also 2 21 21 1h + h = gt + ut - gt2 2

= ut

2

1 22 23 3

u uh h ut ug g

..............(2)

Solving (1) & (2) we get g

uhg

uh94;

92 2

2

2

1

21

7/49/2

2

2

2

1 gugu

hh

Problem : 4.67An object falls from a bridge which is 45 m above thewater. I t falls directly into a small row boat movingwith constant velocity that was 12m from the point ofimpact when the object was released. What was thespeed of the boat ?

Sol. Velocity of boat = sVt

Here,s = 12mt = time of fall of object from bridge

2 2 45= 310

h sg

112V 43

ms

Problem : 4.68Two balls are dropped to the ground from differentheights. One ball is dropped 2s after the other, but bothstrike the ground at the same time 5s after the Ist isdropped.



a) What is the difference in the heights from which theywere dropped ?b) From what height was the first ball dropped?

Sol. a) For the first ball s = h1, u = 0, t = 5 s

21 10 5 9.8 52h x = 122.5mFor the second ball s = h2, u = 0, t = 3s

22

1 1 9.8 9 4.9 9 44.12 2

h gt m

Difference in heights

2 1 122.5 44.1 78.4h h h m b) The first ball was dropped from a height of

h1 = 122.5m Problem : 4.69

Drops of water fall at regular intervals from the roof ofa building of height H = 16m, the first drop striking theground at the same moment as the fifth drop detachesfrom the roof. Find the distance between the successivedrops.

Sol.Step - i : Time taken by the first

drop to touch the ground = 2htg

5

4

3

2

1

For h = 16m, 2tg

.

Time interval between two drops is

int ervalt 1 1 1 21 4 4t tn g

Where n = number of drops.Step - ii :Distance between first and second drops

2 2 21 2 int

1 4 32 erval

S S gt = 7m.Distance between second and third drops

2 2 22 3 int

1 3 22 erval

S S gt = 5m.Distance between third and fourth drops

2 2 23 4 int

1 2 12 erval

S S gt = 3mDistance between fourth and fivfth drops.

2 24 5 int

1 1 02 erval

S S gt = 1m.

Problem : 4.70i) How long does it take a brick to reach the ground ifdropped from a height of 65m ? ii) What will be itsvelocity just before it reaches the ground ?

Hint. i) 2htg

ii) 2V gh

Ans. i) 3.64 s. ii) 35.69ms1

Problem : 4.71A helicopter is ascending vertically with a speed of 8.0ms1. At a height of 120 m above the earth, a packat isdropped from a window. How much time does it takefor the package to reach the ground ?

Sol. Hint : 212

h gt ut Ans. t = 5.83s

Problem : 4.72I f an object reaches a maximum vertical height of 23.0m when thrown vertically upward on earth how highwould it travel on the moon where the acceleration dueto gravity is about one sixth that on the earth ? Assumethat initial velocity is the same.

Hint .1Hg

Ans. 138m

Problem : 4.73An elevator ascends with an upward accleration of 0.2m/s2. At the instant its upward speed is 3m/sec a loose bolt5m high from the floor drops from the celing of theelevator. Find the time untill the bolt strikes the floorand the displacement it has fallen

Sol. Intial velocity of bolt relative to the floor of tyhe elevator = 0

acceleration of bolt relative to the floor of the elevator = (9.8 + 0.2) = 10ms2

If t is time of the descent then 215 102

t

t = 1 secondIf s is the displacement then

212

s gt ut . 1.9s m

Problem : 4.74

A balloon is rising vertically upwards with uniformacceleration 15.7 m/s2 .A stone is dropped from it. After4s another stone is dropped from it. Find the distancebetween the two stones 6 s. after the second stone isdropped

Sol. if f is upward acceleration of the balloon then theacceleration of the stones relative to the balloon is (f+g).The initial velocity of each stone with respect to theballoon is zero.Let s1 and s2 be the distances of the two stones from theballoon after 10s and 6 s respectively. Now



211 1s = f + g 10 = 15.7 + 9.8 100 = 25.5502 2

22 1 25.5 6 25.5 182s 1 2 25.5 32 816s s m

Problem : 4.75A body falls freely from a height of 25m (g=10m/s2) after2sec gravity ceases to act Find the time taken by it toreach the ground?

Sol. 1) Distance covered in 2s under gravity

2 211 1 10 2 202 2

s gt m

velocity at the end of 2sV = gt = (10)2 = 20m/s.Now at this instant gravity ceases to act velocity by here after becomes constant.The remaining distance which is 12520=105 m iscovered by the body with constant velocity of 20m/s.Time taken to cover 105 m with constant velocity isgiven by,

1St =V

1105 5.2520

t s

Hence total time taken to reach the ground= 2 + 5.25 = 7.25 s

Problem : 4.76A solid ball of density half that of water falls freelyunder gravity from a height of 19.6 m and then enterswater. Upto what ?depth will the ball go? Howmuch time will it take tocome again to the water surface Negiect air resistanceand vescosity effects in water (g = 9.8 m/s2).

Sol : Velocity at the surface of water

2 2 9.8 19.6 19.6 / v gh m sAcceleration of a body of density db in the liquid me-dium of density is

' 1 1/ 2

b

d dg g g gd d

Using v2-u2 = 2as in the water

0 - (19.6)2 = 2(-g)hh=19.6 m.

Using s = ut + 212

at , in the water when the ball returns tothe surface, s = o

10 19.6 9.82

t a g

t = 4 s.

Probelem : 4.77A parachutist drops freely from an aeroplane for 10seconds before the parachute opens out. Then he de-scends with a net retardation of 2 m/sec2. His velocitywhen he reaches the ground is 8 m/sec. Find the heightat which he gets out of the aeroplane ?

Sol : Distance he falls before the parachute opens

is 1 100 4902

g m

Then his velocity = gt=98.0 m/s = uVelocity on reaching ground = 8 =

retardation = 22 2 2 u as

228 98 2 2 S

106 90 23854

S m

Total distance = 2385 + 490 =2875 m = height of aeroplane

Problem : 4.78A stone is dropped into a well and the sound of splash isheard after 5.3 sec. I f the water is at a depth of 122.5 mfrom the ground, the velocity of sound in air is

Sol : If t1 is the time taken by stone to reach the ground and t2the time taken by sound to go up, then t1 + t2 = 5.33

Since s = ut +12 at

2

122.5 = 2110 9.82

t t

21

245 2450 259.8 98

t

1 5 t s

2 0.33 t s

Velocity of sound =displacement

time122.5 367 /0.33

m s

Problem : 4. 79A body is thrown vertical ly up with a velocity of100 m/s and another one is thrown 4 sec after the firstone. How long after the first one is thrown will theymeet?

Sol : Let them meet after t sec.

21

11002

S t gt and 221100 4 42

S t g t



221 1100 100 4 42 2

t gt t g t

221400 42

g t t = 1 .4 2 42

g t

28002 4 20, 10 /4

t if g m sg

12 sec t Problem : 4. 80

A lead ball is dropped into a lake from a diving board20 m above the water. I t hits the water with a certainvelocity and then sinks to the bottom of the lake with thesame velocity 6 sec after it is dropped. [g = 10 m/s2].Find the depth of the lake.Ans : 80 m

Sol : Velocity on reaching water 12 10 20 20 ms

Time taken to reach water =2 2 20 2sec

10

h

g depth of water (s) = vt = 20 (6-2) = 80 m

Problem : 4. 81A particle is dropped from point A at a certain heightfrom ground . I t falls freely and passes through threepoints B,Cand D with BC=CD . The time taken by theparticle to move from B to C is 2 seconds and from C to Dis 1 second . Find the time taken to move from Ato B ?

Sol : Let AB=y:BC=CD=h and tAB=t

then 212

y gt A

y

Bh

Ch

D

21 22

y h g t

and 212 32y h g t solving these three equations , we get t=0.5 s

Problem : 4. 82A ball is thrown vertically upward with a velocity 'u'from the balloon descending with velocity v. After whattime, the ball will pass by the balloon ?

Sol : 212r r r

S u t a t

212

O v u gt

2 v ut

g

Problem : 4.83A ball dropped from the 9th storey of a multi - storeyedbuilding reaches the ground in 3 second. In the firstsecond of its free fall, it passes through n storeys, wheren is equal to ( Take g = 10 m s-2)

Sol : 19 10 3 32

y or y = 5 m

Again , 15 10 1 1 52

n or n = 1

Problem : 4.84A stone is dropped into water from a bridge 44.1 mabove the water. Another stone is thrown vertically down-ward 1 s later. Both strike the water simultaneously.What was the initial speed of the second stone ?

Sol: 2 44.1 9 3 ,

9.8t s s s

44.1 = 12 9.8 2 22

v

or 2v = 44.1 - 4.9 4 = 24.5

or 124.5 12.25

2v m s ms -1

Problem : 4.85A ball is dropped from the top of a building. I t takes0.5s to fall past the 3m length of a window some distancefrom the top of the building. I f the velocity of the ball atthe top and at the bottom of the window are VT and VBrespectively thenVT +VB = ?

Sol. 2u vS t

VT

VB

3m3 0.52

T Bv v

12 /T BV V m s

Problem : 4.86Two bal ls are projected vertical ly upwards withvelocitees u1 and u2 from the ground with a time gap ofn seconds. Find the time after which they meet

Sol. If they meet at a height h then 21 21h u t gt u t n2

21 g t n2

21 2 2

1u t gt u t u n2

1 g2

2 21t gn2

gtn

22

1u n gn2t

u gn

b) u ntg 2

if u1 = u2



Problem : 4.87A balloon starts from rest from the ground and moveswith uniform acceleration g/8. When it reaches a heighth a ball is dropped from it the time taken by the ball toreach the ground is

Sol : 2

8 2ghghv

212

h vt gt

v

h

u=0

h

212 2gh

h t gt

21 02 2

ghgt t h

Simplifing and taking only the positive value asnegative value of t is not acceptable we get

2 htg

Problem : 4.88A boy sees a ball go up and then down through a window2.45m high. I f the total time that ball is in sight in 1s,the height above the window the bal l r i ses i sapproximately

Sol : Time during upward crossing of 2.45m

= time during downwad crossing = 12

s

212

h ut gt

1 1 12.45 . 9.82 2 4

u

2

2uH

g

u12

t s

9.8 /g m s

hu = 4.9 2.45 = 2.45

2 2.45 2.452 2 9.8

uHg

0.3H m

4.29 PROJECTILEAny body projected into the air at an angle

other than 900 with the hor izontal near thesur face of the ear th, is called a projectile.

The science of projectile motion is called ballisticsExamples for projectiles :i) A cricket ball thrown by a fielderii) A bullet fired from a gun

iii) A javelin thrown by an athleteiv) A jet of water from a rubber tube impelled into

airNote 4.35 : In two dimensional motion, the positionof a particle is represented by the position vector

r xi yj

V Vx yd dx dyvelocity V xi yj i i jdt dt dtj

x yV and V are components of velocity andrepresent speeds along x and y directions. These twocomponent speeds are independent of each other.

The acceleration vector is given by

jaiajdt

dvi

dtdv

jViVdtd

dtvda yx

yxyx

Again the two components of acceleration areindependent of each other4.30 THE TRAJECTORY OF

PROJECTILE IS A PARABOLA(in absence of air resistance)

Let a body be projected at O with an initialvelocity u that makes an angle with theXaxis.

This velocity can be written as cos sinx yu u i u j u i u jq q .

Due to the fact that two dimensional motion canbe treated as two independent rectilinear motions,the projectile motion can be broken up into twoseparate straight line motions.

i) horizontal motion with zero acceleration.[i.e.,constant velocity as there is no force in horizontaldirection]

ii) vertical motion with constant downwardacceleration = g ( it is moving under gravity)

Y

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