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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

KINEMATICS

SYLLABUS :

Motion in one and two dimension, Projectile motion, Relative Velocity & Circular Motion.

POSITION VECTOR

If the coordinates of a particle are given by (x2, y2, z2) its position vector with

respect to (x1,y1,z1) is given byr

= (x2 - x1) i + (y2 - y1) j +(z2 - z1) k.

Usually, position vector with respect to the origin (0,0,0) is specified and is

given by r

=x i +y j+zk

DISPLACEMENT

Displacement is a vector quantity. It is the shortestdistance between the final and initial positions of a

particle. If1

r is the initial position vector and 2

r

is the final position vector, the displacement vector

is given by12

= rrr .

The magnitude of the displacement is given by

( ) ( ) ( )2122

122

12 zzyyxx ++This is nothing but the straight line distance between two points (x1,y1,z1) and

(x2,y2,z2). The displacement is independent of the path taken by the particle inmoving from (x1 , y1, z1) to (x2, y2, z2)

DISTANCE :If a particle moves along a curve, theactual length of the path is the distance.Distance is always more than or equal todisplacement.

Illustration 1 :

A car travels along a circular path of radius (50 / )m with a speed of 10 m/s. Find

its displacement and distance after 17.5 sec.

Solution :Distance = (speed) time = 10 (17.5) = 175 m

Perimeter of the circular path = 2 (50/ ) = 100 m

The car covers 14

3rounds of the path

If the car starts from A, it reaches B and the displacementis the shortest distance between A and B

Displacement = 22 RR + =

= 2502 R m.

)z,y,x(P 111)z,y,x(Q 222

A

B

1

r

r

2

r

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657INSTANTANEOUS AND AVERAGE VELOCITY :

Ifr is the displacement of the particle in time t, the average velocity is

given by

V average =

t

r

tt

rr

=

12

12

= Find value - Initial value. The above definition is valid for any magnitude

of large or small. But when is infinitesimally small, the instantaneousvelocity is obtained.

V instantaneous =

Ltt 0

t

r

=dt

rd

In normal notation, velocity refers to the instantaneous velocity.

SPEED :

Speed =time

cetanDis

When the time under consideration is very small, distance becomes equal tothe displacement and speed becomes the magnitude of instantaneousvelocity. speed is represented only by its magnitude where as velocity isrepresented by magnitude as well as direction.

INSTANTANEOUS AND AVERAGE ACCELERATION :

If

V is the change in velocity in time t, average acceleration is given by

averagea

=t

V

.

When becomes infinitesimally small,t

VLtt

0=

dt

Vd

which gives the instantaneous acceleration.In normal notation, acceleration refers to the instantaneous acceleration.

==

dt

rd

dt

d

dt

Vda =

2

2

dt

rd

It may be noted here that magnitude of2

2

dt

rd is not equal to2

2

dt

rdalways (as in

the case of circular motion)

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Illustration 2 :

A bus shuttles between two places connected by a straight road with uniform speedof 36 kmph. If it stops at each place for 15 minutes and the distance between thetwo places is 60 km, find the average values of (a) Speed

(b) Velocity(c) acceleration between t = 0 and t = 2 hours and the instantaneous values of(d) Velocity (e) acceleration at t = 2 hrs.Solution :

Time taken for forward trip =36

60=3

5hrs.

Time of stoppage = 15 min = 0.25 hrs.Time available for return trip = 2 - 5/3 - 0.25 = 1/12 hrs.Distance travelled in the return trip = (36) 1/12 = 3 km.

a) Average speed =timeTotal

distanceTotal=

2

360+= 31.5 kmph.

b) Average velocity =Time

ntDisplaceme=

2

360= 28.5 kmph

c) Average acceleration =Time

velocityinchange

=t

VV

12 =( ) ( )

2

3636 += - 36 km/H2 = -

360

1m/s2

d) Velocity at t = 2 hours = - 36 kmphe) Acceleration at t = 2hours = 0 as there is no change in velocity

Illustration 3 :A car travels towards North for 10 minutes with a velocity of 60 Kmph, turnstowards East and travels for 15 minutes with a velocity of 80 kmph and then turnstowards North East and travels for 5 minutes with a velocity of 60 kmph. For thetotal trip, find a) distance travelled b) displacement c) average speed d)average velocity and e) average acceleration.

Solution :Total time taken = (10 + 15 + 5)min = 1/2 houra) Distance travelled = d1 + d2 + d3

= 60

60

10+ 80

60

15+ 60

60

5

= 10 + 20 + 5 = 35 km

b) displacement

++= 321 SSSS

= 10 j + 20 i +5 cos 450 i + 5 sin 45 j

= 23.5 i + 13.5 j

Magnitude of displacement = ( ) ( )22 513523 .. + ~ 27 km

kmd 31 =

kmd 60=

1

S

2

S

S

3

S

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c) Average speed =hr

kmdiTotal

=

2

1

35

Time

travelledstance

= 70 kmph.

d) Average velocity =Time

ntDisplaceme=

( )jij.i. 2 74 7

2

1

51 352 3+=

+

kmph.

at an angle with the East given by Tan =523

513

.

.

Magnitude of average velocity = 22 2747 + = 54 kmph

e) Average acceleration =Time

yin velocitchange

=Time

velocityInitialvelocityFinal

=

( )2

6 04 56 04 56 00 jjs i nic o s +

= ( )ji 92 1 km/H2

at an angle with the East given by Tan =21

9

Magnitude of average acceleration = 22 921 + ~ 23 km/H2 ~ 1.8 x 10-3

m/s2

Illustration 4 :

A car moving along a circular path of radius R with uniform speed covers an angle during a given time. Find its average velocity and average acceleration during thistime.

Solution :

Let V be the speed of the car

V =time

Distance=

t

Rwhere is in radians.

Displacement= + cosRRR 222 2 from the triangleOAB

= 2R sin /2

Average velocity =Time

Diplacemnt=

V

R

sinR2

2

=

2

2 sinV

Average acceleration =t

V

time

velocityinChange =

A V

B

Vo

V

V

V

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0 - (22.5)2 = 2(-10) d1 d1 =( )

20

5222

.= 25.3 m

d2 can be found from S = ut + 1/2 at2 applied along the down warddirection starting from the top most pointd2 = 0 (t - t0) + 1/2 g (t - t0)2 = 1/2 (10) (4 - 2.25)2 = 15.3 m

Distance travelled in 4 sec = 25.3 + 15.3 = 40.6 m

Displacement in 4 sec = d1 - d2 = 25.3 - 15.3 = 10 mDisplacement can also be found directly by applying S = ut + 1/2at2 along the verticalDisplacement in 4 sec = 22.5 (4) - 1/2 (10) (4)2 = 10M

d) 3rd second is from t = 2 sec to t = 3 sec.

Displacement in the 3rd second = u +2

a(2n - 1)

= 22.5 -2

10(6 - 1) = -2.5 m

When there is no change in the direction of the motion along a straight line,distance will be equal to displacement. When the particle reverses its

direction during the time under consideration, distance will be more than thedisplacement and the time at which the reversal is taking place must befound.

When the particle reverses its direction, its velocity becomes zero.

using V = u + at, 0 = 22.5 - 10 (t0) t0 = 2.25 secd = d1 + d2

using the formula S = ut + 1/2 at2

d1 = [22.5 (2.25) - 1/2 (10) (2.25)2 ] - [22.5 (2) - 1/2 (10)(2)2]

= 0.31 m

Along the downwards vertical starting from the topd2 = 0 (3 - 2.25) + 1/2 (10) (3 - 2.25)2 = 2.81 m

d = 0.31 + 2.81 = 3 .12 m

KINEMATICAL EQUATIONS ( VARIABLE ACCELERATION ) :

When the acceleration is variable, the kinematical equation take the form

V =dt

dxa =

2

2

dt

xd

dt

dV =

a = dx

VdV

dt

dx

dx

dV

=

V = u +

t

dta0

x = ut + dtdtatt

00

and V2 - u2 = 2 x

dxa

0

Illustration 6 :The position coordinate of a particle moving along a straight line is given by x = 4t3-3t2+4t+5. Find a) Velocity and acceleration as a function of time b)Displacement as a function of time c) the time at which velocity becomes zero andthe acceleration at this time d) the time at which acceleration becomes zero and

the velocity at this time.

sect 3=sect 2=

sec.t 252=

1d 2d

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657Solution :

a) V =dt

dS=

( )

dt

xxd 0 =dt

dxWhere x0 is the initial position coordinate

which is a constant

=dt

d(4t3 - 3t2 + 4t + 5) = 12 t2 - 6t + 4

a = dtdv = dtd (12t2 - 6t + 4) = 24t - 6

b) Displacement = (position coordinate at time t) - (position coordinate at t= 0)

= (4t3 - 3t2 + 4t + 5) - (5)

= 4t3 - 3t2 + 4t

c) When V = 0, 12 t2 - 6t + 4 = 0 t =12

4893

since this value is imaginary, the velocity never becomes zero.

d) When a = 0, 24t 6 =0 and t = 24

6

= 4

1

units and the velocity of

the particle at this time, V= 124

134

4

16

4

12

=+

units

Illustration 7 :The velocity of a particle moving in the positive direction of the x axis varies as V =

x where is a positive constant. Assuming that at the moment t = 0 theparticle was located at the point x = 0, find a) the time dependence of the velocityand the acceleration of the particle b) the mean velocity of the particle averagedover the time that the particle takes to cover the first S meters of the path.

Solution :

a) V =dt

dx= x =

tx

dtx

dx

00

2 x = t and x =4

22t

V =2

2tand a =

dt

dV=

2

2

b) Mean velocity =

time

ntDisplaceme

Displacement = S, and the time taken for this displacement t =S2

Mean velocity =

( )

S

S

2 =2

S

Mean velocity can also be found from the following formulae

V mean =

dx

dxV

when V is a function of x

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and V mean =

dt

dtV

when V is a function of time

KINEMATICAL EQUATIONS IN VECTOR FORM ( CONSTANT ACCELERATION ) :

tauV

+=

= S.au.uV.V 2

2

2

1tatuS

+=

The above equations are useful in 2 and 3 dimensional motion.

Illustration 8 :A particle moving on a horizontal plane has velocity and acceleration as shown inthe diagram at time t = 0. Find the velocity and displacement at time 't'.

Solution :METHOD - I

u = u cos30

0 i + u cos 60 j =2

3u i +

ju

2

a = - a cos 45 i - a cos45 j = 2

a i -

2

a j

V = tau + = it

au

22

3 + jtau

22

The magnitude of the velocity =22

2222

3

+

t

auatu

2

2

1tatuS

+= = jt

at

uit

aut

+

22

22

1

222

1

2

3= Sx i + Sy j

The magnitude of the displacement = 22 ySSx +

030

045a

x

y

u

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657METHOD - IIThis can be solved by vector addition method also. It may be

noted here thatu t will be along the direction of

u , ta

and

2

2

1ta

will be along the direction of

a

tauV

+=

Since the angle between u and ta

is 1650, the magnitude of

the velocity is

( ) ( ) 022 1652 cosatuatu ++

2

2

1tatuS

+=

Since the angle betweenu t and

2

2

1ta

is 1650, the

magnitude of the displacement is

( ) ( )

022

22

1652

1

22

1

cosatutatut

+

+

KINEMATICAL EQUATIONS IN RELATIVE FORM ( CONSTANT ACCELERATION ) :

When two particles A and B move simultaneously with initial velocitiesu A

andBu

, at any time 't'

= AAB VV -

BV

;

= BAAB SSS ;

= BAAB aaa

ABU=

BA UU

ABV = tau ABAB

+2

2

1tatuS ABABAB

+=

whereABX

means parameter X of A with respect to B.

Similarly ifr

is the position coordinate at time 't' and0r

is the initial position

coordinate at time 't' = 0,

2

10 ++=

turr AAA 2taA

2

1

0 ++=

turr

BBB 2

taB

= ABAB rr 0

+ 22

1tatu ABAB

+ gives the position coordinate of A with respect to B

at any time.

ABr gives the distance between A and B at any time 't'.

Illustration 9 :A loose bolt falls from the roof of a lift of height 'h' moving vertically upward withacceleration 'a'. Find the time taken by the bolt to reach the floor of the lift and thevelocity of impact.

030

u

V

ta

045

2

2

1ta

030

tu

S

045

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Solution :

jhS b =

as the bolt travels a distance 'h' down wards before hitting the

floor

= aaa bb = (-g

j ) - (a j ) = - (g + a) j

= uuu bb = ju - ju = 0 as they have the same initial velocity upwards

2

2

1tatuS bbb

+=

- h j = 0 -2

1(a + g) t2 j t =

ga

h

+2

Velocity of impact is nothing but the relative velocity of the bolt with respectto the lift

Vimpact = tauV bbb

+=

= - (a+g) j ga

h

+

2= ( )gah + 2 j

Illustration 10 :Two particles A and B move on a horizontalsurface with constant velocities as shown inthe figure. If the initial distance of separationbetween them is 10 m at t=0, find thedistance between them at t = 2 sec

Solution :

Distance between them = ABr

Taking the origin at the initial position of A

20

2

1taturr ABABABAB

++=

ir AB 100 =

( ) ( )jcosicosjcosicosUUu BAAB 0000 301060104521045210 +==

= 5 i - 18.7 j

0==

BAAB aaa

( ) ( )j.iirAB 718510 += t

At t = 2 sec , j.rAB 437=

and ABr

= 37.4 units

DISPLACEMENT - TIME GRAPHS :The displacement is plotted along 'y' axis and the time along 'x' axis. The

slope of the curvedt

dSgives the instantaneous velocity at that point. The average

slope between two pointst

S

gives the average velocity between these points.

m1 00

6 0

s/mu B 1 0=

04 5

s/muA 21 0=

A B

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657Rate of change of slope gives the acceleration. If the slope is positive and

decreases with time, the particle is under retardation. If the slope is positive andincreases with time, the particle is under acceleration, constant slope implies zeroacceleration.

Illustration 11 :The displacement - time graph of a particle moving

along a straight line is given below. Finda) the time at which the velocity is zerob) the velocity at time t = 1 secc) the average velocity between t = 2 sec and t = 4sec

Solution :a) Velocity is zero when the slopoe is zero which happens at t = 2 secb) Since any point (x,t) lies on the circle of radius 2 m and centre (2,0),

(x-0)2 + (t - 2)2 = 22 x = ( ) 224 t

velocity is given by the slope dt

dx= V

V =dt

d ( )

( )( )( )

=

22

242

124

2

2t

t

t = +

3

1

Since the slope is +ve between t = 0 and t = 2, v =3

1m/s

c) Average velocity =time

ntDisplaceme=

24

2

O

= -1 m/s

VELOCITY - TIME GRAPH :

If velocity is plotted on 'y' axis and time is plotted on x axis, the slope of the

curve at any pointdt

dvgiven instantaneous acceleration. The average slope

between two pointst

v

gives average acceleration.

The total area between the curve and the time axisgives distance where as algebraic sum of the areasgives displacement.

Distance = A1

+ A2

+ A3

Displacement = A1 - A2 + A3The nature of acceleration can be found from the rate of change of slope

Illustration 12 :The velocity time graph of a particle moving along a straight line has the form of aparabolav = (t2 - 6t + 8) m/s . Finda) the distance travelled between t = 0 second t = 3 sec

b) the velocity of the particle when the acceleration is zero

V1A

2A

3A

t

0

m2x

2 4 t

circleSemi

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657c) the acceleration of the particle when the velocity is zerod) the velocity of the particle when the acceleration is zero

Solution :a) Distance = area OAB + area BCF which can be

obtained by the method of integration.Since at the points B and D, velocity becomes

zero t2 6t + 8 = 0 t = 2 sec and 4secSince F is in between B and D, the time

corresponding to F is2

42+= 3 sec. Similarly A

corresponds to t = 0 and E corresponds to t = 6 sec

Area OAB = A1 = ( )

+=+=

2

0

2

0

232

2

0

82

6

386 t

ttdtttdtV =

3

20m

Area BCF = A2 = - mttt

dtV3

28

2

6

3

233

2

=

+=

Distance = m3

22

3

2

3

20=+

b) displacement between t = 3 sec and t = 6 sec = A4 - A3 = A1 - A2 =

3

2

3

20 = 6m

c) a =dt

dv= 2t - 6

When V = 0 ; t = 2sec and 4 sec

a = 2(2) - 6 and 2(4) - 6= - 2 m/sec2 and 2 m/sec2

d) When a = 0, 2t - 6 = 0 and t = 3 secV = 32 - 6(3) + 8 = -1 m/secPROJECTILE MOTION :

At the top most point Vy = 0 and

Vx = u cosFrom Vy = uy + ay t, 0 = usin -

gt

t =g

sinu

Time of flight = 2t =g

sinu 2

From 22 yy uV = 2ay Sy, 0 - (u sin )2 = 2 (-g) H and H =g

sinu

2

22

Range = (Time of flight) (horizontal velocity) =

( )g

sinucosu

g

sinu =

22 2

Range is maximum when = 450 and Rmax =g

u2

R

H

T

u

V

EA

1AO

B F D4A

3A

2A t

C

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H

R=

g

sinu

g

cossinu

2

2

22

2

= 4 cot

The velocity of the particle at any time 't' is given by jViVV yx +=

V = (ucos ) i + (usin - gt) j

The magnitude of the velocity = ( ) ( )22 gtsinucosu +

If is the angle made by the velocity at any time 't' with the horizontal,

Tan =

cosu

gtsinu

Taking the origin at the point of projection, the 'x' and 'y' coordinates at anytime 't' are given by

x = u cos t and y = usin t2

1 gt2

Eliminating 't' from x and y

y = u sin

cosux

2

1g

2

cosux

= x tan -22

2

2 cosu

gxwhich is the equation of a parabola.

It may be noted here that the velocity of the projectile will be alwaystangential to its path. The equations of projectile motion derived above arevalid only for constant acceleration due to gravity 'g'.

Illustration 13 :

A particle is projected from the horizontal at an inclination of 600 with an initialvelocity 20 m/sec.

Assuming g = 10 m/sec2 find a) the time at which the energy becomes threefourths kinetic and one fourth potential b) the angle made by the velocity at thattime with the horizontal c) the x and y coordinates of the particle taking theorigin at the point of projection.

Solution :

a) Let V be the velocity when the given condition is fulfilled2

1mV2 =

4

3(

2

1mu2)

V =2

3u= 10 3 m/sec

( ) jgtsinuicosuV +=

= 20 cos 600 i + ( )tsin 106020 0 j = 10 i + ( jt10310 V = 10 3 102 + ( )210310 t = ( )2310

Solving t = 23 sec

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657t = 23 while rising up and t = 23+ while coming down

b) Tan =

cosu

gtsinu

=( )

10

2310310 = + 2

c) x = ucos t

= 10 23 = 10 ( ( morm 231023 +and y = u sin t - 2

1

gt2

= 10 3 23 - 5 ( ) m523 2 =

PROJECTILE MOTION ON AN INCLINED PLANE :

Let be the inclination of the plane and

the particle is projected at an angle with theinclined plane. It is convenient to take thereference frame with x' along the plane and y'

perpendicular to the plane. gcos will bethe component of the acceleration along the

downward perpendicular to the plane and g sin will be the component of theacceleration along the downward direction of the inclined plane.

Along the plane, the kinematical equations take the formtauV 'x'x'x +=

tsingcosuV 'x =

22

1tat'uS 'xx'x +=

'xS

= ucos t - 2

1

gsin t

2

2

'xV - 'x'x'x Sau 22 = 2'xV - (ucos )

2 = 2 (-g sin ) 'xS

Similarly perpendicular to the plane, the kinematical equations take the formtauV 'y'y'y += tcosgsinuV 'y =

2

2

1tatuS 'y'y'y +=

2

2

1tcosgtsinuS 'y =

'y'y'y'y SauV 222 = ( ) ( ) 'y'y ScosgsinuV = 2

22

Here it may be noted that,When the particle strikes the inclined plane 'yS = 0

When the particle strikes the inclined plane perpendicular to it, 0='yS and0='xV

When particle strikes the inclined plane horizontally 0='yS and yV = 0

Illustration 14 :From the foot of an inclined plane of

inclination , a projectile is shot at an angle with the inclined plane. Find the relation

between and if the projectile strikes theinclined plane

a) perpendicular to the plane

'y

u 'x

'y y

u 'x

co sgsing

x

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657b) horizontally

Solution :

a) Since the particle strkes the plane perpendicularly 0='yS and 0='xV

u sin t -2

1g cos t2 = 0 and u cos - g sin t = 0

t =

cosg

sinu2

and t =

sing

cosu

=

sing

cosu

cosg

sinu2 2 Tan = cot

b) Since the particle strikes the plane horizontally 'yS = 0 and 0=yV

u sin t -2

1g cos t2 = 0 and u sin ( + ) - gt = 0

t =

cosg

sinu2=

( )g

sinu +

( )g

sinu

cosg

sinu +=2

cos

sin2= sin

( + )

CIRCULAR MOTION :

When a particle moves in a circle of radius R with constant speed V, its calleduniform circular motion.

When the particle covers , the direction of velocity also changes by

without change in magnitude. Change in velocity V will be towards the centre ofcurvature of the circular path which causes centripetal acceleration. is called theangular position (or) angular displacement.

Centripetal acceleration, ra =t

V

The rate of change of angular position is known as angular velocity ( )

Time period of circular motion T =V

R2

In the same time the particle covers an angle 2 from which angular velocity canbe found as

=T

2=

R

V

R

V =

2

2

t =

=V

Rand V = + cosVVV 222 2 = 2V sin

2

When is small sin ~ V = V

centripetal acceleration =t

V

=

( )

V

R

V

=R

V2

V

V

V

V

V

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657When speed of the particle continuously changes with time, the

tangential acceleration is given by ta =dt

dV

The rate of change of angular velocity is called the angular acceleration ( )since ra and ta are perpendicular to each other, the resultant

acceleration is given by a = 22 tr aa +

Angle made by the resultant with radius vector Tan =r

t

a

a

Illustration 15 :The speed of a particle in circular motion of radius R is given by V = Rt2. Find thetime at which the radial and the tangential accelerations are equal and the distancetraveled by the particle upto that moment.

Solution :

tr aa =

dt

dV

R

V=

2

= 2Rt t = 31

2

Distance travelled = 3

1

2

0

dtV =3

1

23

3

Rt =3

2R

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657RADIUS OF CURVATURE

When a particle is moving in a planeR

Va r

2

= where V is the instantaneous

velocity and R is the radius of curvature at that point.

Radius of curvature =ra

V2

If the path of the particle is given by y = f(x), radius of curvature can also be

found from the formula R =

2

2

2

32

1

dx

yd

dx

dy

+

Illustration 16 :

A particle is projected with initial velocity 'u' at angle with the horizontal. Findthe radius of curvature at a) point of projection b) the top most point

Solution :

a) at the point of projection P, V = u and ra = g cos

R =ra

V2

=cosg

u 2

b) at the topmost point T, V = ucos and ga r =

R =ra

V2

=g

cosu 22

SHORTEST DISTANCE OF APPROACH :

When two particles A and B are moving simultaneously, their positioncoordinates at any time 't' are given by (when the accelerations are uniform)

2

02

1taturr AAAA

++= and 20

2

1taturr BBBB

++=

The distance between them at any time 't', S = ABr

Where 202

1taturr ABABABAB

++=

The distance between them becomes minimum whendt

dS=0 from which the

time at which it becomes minimum can be found. Substituting the value of

time so obtained inABr

, S min can be found.

Illustration 17 :Two ships A and B move with constantvelocities as shown in the figure. Find theclosest distance of approach betweenthem

u

p

T

A

km10

o3 0

North

kmphVA 20=

kmphVB21 0

=

East

045

O km20 B

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Solution :

jrA 100 =

ir B 200 =

jcosicosVA 302060200 = jcosicosVB 4521045210 +=ji 31010 = = 10 i + 10 j

0=

Aa 0=

Ba

tVrr AAA

+= 0 tVrr BBB

+= 0= 10 t i + ( ) jt31010 = (20 + 10 t) i + 10 t j

BAA B rrr

= = - 20 i + ( jtt 1031010

S = ABr

= ( ) ( )22 103101020 tt +

When the distance between A and B is minimumdtdS = 0

( ) ( )

+22

1031010202

1

tt ( (( 1031010310102 tt = 0

10 - 10 3 t - 10 t = 0 t =

+ 311

hr

Substituting this value of time in the expression for S,S min = 20 km

CYCLIC MOVEMENT OF PARTICLES :

When three or more particles located at the vertices of a polygon of side lmove with constant speed V such that particle 1 moves always towards particle 2and particle 2 moves always towards 3 particle etc., they meet at the centre of thepolygon following identical curved paths.

Time of meeting =approachVelocityof

seperationInitial

Velocity of approach is the component of the relative velocity along the linejoining the particles.

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Illustration 18 :

Six particles located at the six vertices of a hexagon of side l move with constantspeeds V such that each particle always targets the particle in front if it. Find thetime of meeting and the distance travelled by each particle before they meet

Solution :

t =approachofVelocity

seperationInitial

= 060cosVV

=V

2

Since they move with constant speed V, the distance

travelled by each particle in time t =V

2is d = Vt =

VV

2= 2 l

RIVER PROBLEMS :

IfrV

is the velocity of the river andbV

is the velocity of the boat with respect tostill water, the resultant velocity of the

boatRV

=

rb VV

+Only the perpendicular component of theresultant velocity helps in crossing the

river. Time of crossing, t = cosVw

bwhere

'w' is the width of the river.

The boat crosses the river in the least time when = 0The parallel component of the resultant velocity determines the drift.Drift is the displacement of the boat parallel to the river by the time the boatcrosses the river

Drift , x = ( ) sinVV br

cosV

w

b

Zero drift is possible only when Vr = Vb sin . When Vr > Vb zero drift is notpossible.

Illustration 19 :

A river of width 100 m is flowing towards East with a velocity of 5 m/s. A boat whichcan move with a speed of 20 m/s with respect to still water starts from a point onthe South bank to reach a directly opposite point on the North bank. If a wind isblowing towards North East with a velocity of 5 2 m/s, find the time of crossingand the angle at which the boat must be rowed.

Solution :

jcosisinVb +=

2020

rV

= 5 i

V

V

V

V

V

06 0

B

rV

A

C

bV

rVwV

45

bV

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657V w = 5 2 cos 45 i + 5 2 cos 45 j = 5 i + 5 j

RV

= Resultant velocity of the boat =bV

+

rV

+wV

= ( - 20 sin + 5 + 5) i + (20 cos + 5) j

For reaching directly opposite point, the component of the resultant velocity

parallel to the river must be zero

- 20 sin + 10 = 0 sin =2

1and = 300

Since time of crossing depends only on the perpendicular component of theresultant velocity.

t =520 +cos

w=

53020

100

0 +cos = 4.48 sec

WORKED OUT OBJECTIVE PROBLEMS

EXAMPLE : 01A point moves along 'x' axis. Its position at time 't' is given by x2 = t 2 + 1.Its acceleration at time 't' is

A)3

1

xB)

2

11

xx C)

2x

tD)

3

2

x

t

Solution :

x = 12

+t ; dt

dx

= 12

1

2 +t (2t) = 12 +t

t

a =2

2

xt

xd=

22

2

2

1

2

12

1

+

+

+

t

)t(

t

tt

= ( )32 11

+t=

3

1

x

EXAMPLE : 02

A body thrown vertically up from the ground passes the height 10.2mtwice at an interval of 10 sec. Its initial velocity was (g = 10 m/s2)A) 52 m/s B) 26 m/s C) 35 m/s D) 60 m/s

Solution :Displacement is same in both casess = ut + 1/2 at2

10.2 = ut -2

1(10) t2 t =

10

2042 uu

t1 =10

2042 uu and t2 =10

2042 + uu t = t2 - t1 = 10 sec

2042 u = 50 u2 = 2500 + 204 u = 52 m/s

EXAMPLE : 03

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657A car starts from rest moving along a line, first with acceleration a= 2m/s2, then uniformly and finally decelerating at the same rate and comesto rest. The total time of motion is 10 sec. The average speed during thistime is 3.2 m/s. How long does the car move uniformlyA) 4 sec B) 6 sec C) 5 sec D) 3 sec

Solution :

Let the car accelerate for time 't' and move uniformly with v = at for time t1Since the magnitudes of acceleration and deceleration are same, the time ofdeceleration is also 't'.

t + t 1 + t = 10 sec

Average speed =time

Distance=

( )

10

2

1

2

1 21

2

++

attatat= 3.2

2t2 + 2tt1 = 32 2 322

102

2

101

1

2

1 =

+

t

tt

Solving t1 = 6 sec This problem can be solved using velocity time graph also.

EXAMPLE : 04

A particle has an initial velocity of ( )ji 43 + m/s and a constantacceleration ( )ji 34 m/s2. Its speed after 1 sec will be equal toA) zero B) 10 m/s C) 5 2 m/s D) 25 m/s

Solution :

tauV

+= = ( ) ( )jiji 3443 ++ (1) = 7 i - j

Speed = magnitude ofV =

22 17 + = 5 2 m/s

EXAMPLE : 05

An aeroplane flies along a straight line from A to B with air speed V andback again with the same air speed. If the distance between A and B is land a steady wind blows perpendicular to AB with speed u, the total timetaken for the round trip is

A)V

2B) 22

2

uV +

C) 222

uV

V

D) 222

uV

Solution :The resultant velocity of the plane must be along AB duringforward journey.

t1 = 22 uVVR =

A

RV

B

V

u

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657During return journey, the resultant velocity of the plane mustbe along BA

t2 = 22uVVR

=

Total time t = t1 + t2 = 222

uV

EXAMPLE : 06

A particle is thrown with a speed 'u' at an angle with the horizontal.

When the particle makes an angle with the horizontal its speed changesto V. Then

A) V = u cos B) V = ucos cos C) V = u cos sec

D) V = usec cos

Solution :Since the horizontal component of the velocity of a projectile always remains

constantu cos =V cos V=ucos sec

EXAMPLE : 07

Two shells are fired from a cannon with same speed at angle andrespectively with the horizontal. The time interval between the shots is T.They collide in mid air after time 't' from the first shot. Which of thefollowing conditions must be satisfied.

A) > B) t cos = (t -T) cos

C) (t-T) cos =cos D) (usin )t -2

1gt2 =(usin )

(t-T)-2

1g(t-T)2

Solution :When they collide, their 'x' and 'y' components must be same

ucos t = u cos (t-T) cos t = cos (t-T)

(usin ) t -2

1gt2 = (usin ) (t-T) -

2

1g (t-T)2

Since cos = cos

t

T1 and T < t

cos < cos and >

EXAMPLE : 08A particle is projected from a point 'p' with velocity 5 2 m/s perpendicularto the surface hollow right angle cone whose axis is vertical. It collides atpoint Q normally on the inner surface. The time of flight of the particle isA) 1 sec B) 2 sec C) 2 2 sec D) 2 sec

Solution :

RV

V

A

Bu

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It can be seen from the diagram thatV becomes perpendicular to

u .

u = ucos45

0 i + u sin45 j

V =

u +

a t = (ucos 45 i + usin45 j ) - (gt) j

WhenV becomes perpendicular to

u ,

V .

u = 0

u2 cos2 45 + u2 sin2 45 - (usin45) gt = 0 t = 45singu

= 1 sec

EXAMPLE : 09A man walking Eastward at 5 m/s observes that the wind is blowing fromthe North. On doubling his speed Eastward he observes that the wind isblowing from North East. The velocity of the wind isA) (5i+5j) m/s B) (5i - 5j) m/s C) (-5i +5j) m/s D) (-5i - 5j) m/s

Solution :

let jViVVw 21 +=

In the first casemwwm VVV

= = ( )jViV 21 + - ( )i5

Since no component along East is observed V1 - 5 = 0 V1 = 5 m/sIn the second case

mwwm VVV

= = (V1 i + V2j ) - (10 i ) = ( ) jViV 21 10 +

Since the wind is observed from North East the components along North and

East must be sameV1 - 10 = V2 V2 = - 5 m/sV w = (5i - 5j) m/s

EXAMPLE :10From a lift moving upward with uniform acceleration 'a', a man throws aball vertically upwards with a velocity V relative to the lift. The time afterwhich it comes back to the man is

A)ag

V

2

B)ag

V

+ C) agV

+2

D) 222

ag

Vg

Solution :

Since the velocity of the ball is given relative to the liftblV

= V j

When the ball comes back to the man, its displacement relative to the lift is

zeroblS

= 0

bla

= lb aa

= (-g) j - a j = - (g + a) j

Applying S = ut + 1/2 at2 in relative form

blbl VS

= t + 21

bla

t2

0 = ( )jVt +2

1 ( )( )jag + t2 t =

ag

V

+2

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 0406460665710. A particle has an initial velocity of 9m/s due east and a constant acceleration

of 2m/s2 due west. The distance covered by the particle in the fifth second ofits motion isA) 0 B) 0.5m C) 2m D) none of these

11. Two particles are projected simultaneously in the same vertical plane from the

same point, with different speeds u1 and u2, making angles 1 and 2respectively with the horizontal , such that u1 cos 1 = u2cos 2. The pathfollowed by one, as seen by the other (as long as both are in flight), is :A) a horizontal straight line B) a vertical straight line C) a parabola

D) a straight line making an angle | 1 - 2| with the horizontal12. A particle starts from the origin of coordinates at time t = 0 and moves in the

xy plane with a constant acceleration in the y-direction. Its equation of

motion is y = x2. Its velocity component in the x-direction is

A) variable B)2

C)2

D)2

13. A train starts from station A with uniform acceleration for some distance

and then goes with uniform retardation for some more distance to come to

rest at station B. The distance between station A and B is 4 km and the traintakes 4 minute to complete this journey. If and are in km (min)-2

then

A) 211=

+

B)4

11=

+

C) 2111

=

+ D) 4

111=

+

14. The driver of a train moving with a speed v1 sights another train at a distance

d, ahead of him moving in the same direction with a slower speed v2. Heapplies the breaks and gives a constant de-acceleration 'a' to his train. For nocollision, d is

A) =( )

a2

vv2

21 B) >( )

a2

vv2

21 C) t2

B) R' < R, H' < H, '1t > t1 and'2t < t2

C) R' < R, H' > H, '1t > t1 and'2t < t2

D) R' < R, H' < H, '1t < t1 and'2t > t2

35. Speed of a particle moving in a circle varies with time as, v = 2t. Then :

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657A) angle between velocity vector and acceleration vector is increasing withtime.B) a is constant while ar is increasing with time.C) both A and B are correct D) both A and B are wrong.

36. Initial velocity and acceleration of two particles are as shown in fig. Assumingthe shown direction as the positive, vBA versus time graph is as :

A) B) C) D)37. A graph is plotted between velocity (v) and displacement (s) of particle

moving in a straight line. Here v is plotted along y-axis and 's' along x-axis.Choose the correct option.A) slope of this graph at any point always gives us the ratio of velocity anddisplacement at that point.7

B) slope represents a/v under all the conditions. (a = acceleration)C) both A and B are correct D) both A and B are wrong.

38. In a projectile motion if a person wants to increase the maximum height to 2times but simultaneously want to decrease the range same number of time.He can achieve it by increasing tan of angle of projectionby ....... times.A) 2 B) 4 C) 3 D) 2

39. The velocity of a particle moving in a straight line varieswith time in such a manner that v versus t graph isrepresented by one half of an ellipse. The maximum

velocity is m and total time of motion is t0

i) Average velocity of particle is /4 m ii) Such motion can notbe realized in practical termsA) Only (i) is correct B) Only (ii) is correctC) Both (i) and (ii) are correct D) Both (i) and (ii) are wrong

40. Starting from rest, a particle rotates in a circle of radius R = 2 m with an

angular acceleration = /4 rad/ s2. The magnitude of average velocity of theparticle over the time it rotates quarter circle isA) 1.5 m/s B) 2 m/s C) 1 m/s D) 1.25m/s

41. In a car race car A takes t0 time less to finish than car B and passes thefinishing point with a velocity v0 more than car B. The cars start from rest and

travel with constant accelerations a1 and a2. Then the ratio 0/t0 is equal to

A)2

21

a

aB)

2

aa 21+ C) 21 aa D)1

22

a

a

42. A rod of length 1 leans by its upper end against a smoothvertical wall, while its other end leans against the floor. Theend that leans against the wall moves uniformly downward.ThenA) The other end also moves uniformlyB) The speed of other end goes on decreasingC) The speed of other end goes on increasingD) The speed of other end first decreases and then increases

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 0406460665743. A particle is moving along a circular path of radius 5 m and with uniform speed

5 m/s. What will be the average acceleration when the particle completes halfrevolution?

A) zero B) m/s2 C) 10 m/s2 D) 10/ m/s2

44. The velocity displacement graph of a particle moving along astraight line is shownThe most suitable acceleration-displacement graph will be

A) B)

C) D)

1 2 3 4 5 6 7 8 910

11

12

13

14

15

16

17

18

19

20

C B C A C A B C C B B D A B B A A A B B21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

B C A C C B A C A A C D B D C A B B C C41

42

43

44

C A

LEVEL II1. A particle starts from rest at time t = 0 and moves

on a straight line with an acceleration whichvaries with time as shown in fig. The speed of theparticle will be maximum after how many seconds

A) 4s B) 6sC) 8s D) 10s

2. Due to air a falling body faces a resistive force proportional to square ofvelocity v, consequently its effective downward acceleration is reduced and isgiven by a = g - kv2 where k = 0.002m-1. The terminal velocity of the fallingbody isA) 49m/s B) 70m/s C) 9.8m/s D) 98m/s

3. A balloon is rising with a constant acceleration of 2m/s2. At a certain instantwhen the balloon was moving with a velocity of 4m/s, a stone was droppedfrom it in a region where g = 10m/s2. The velocity and acceleration of stone asit comes out from the balloon are respectively.A) 0, 10m/s2 B) 4m/s, 8m/s2 C) 4m/s, 12m/s2 D) 4m/s, 10m/s2

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 040646066574. A stone is thrown vertically up from the top of a tower with some initial

velocity and it arrives on the ground after t1 seconds. Now if the same stone isthrown vertically down from the top of the same tower with the same initialvelocity, it arrives on ground after t2 seconds. How much time will the stonetake to reach the ground if it is dropped from the same tower ?

A)2

tt 21 + B)2

tt 21 C) 21 tt + D) 21tt

5. Two particles A and B start from the same point and slide down throughstraight smooth planes inclined at 300 and 600 to the vertical and in the samevertical plane and on the same side of vertical drawn from the starting point.The acceleration of B with respect to A isA) g/2 in vertical direction B) g 2/3 at 450 to verticalC) g/ 3 at 600 to vertical D) g in vertical direction

6. A particle starts from rest at the origin and moves along X-axis withacceleration a = 12-2t. The time after which the particle arrives at the originisA) 6s B) 18s C) 12s D) 4s

7. Figure represents position (x) versus time (t) graphfor the motion of a particle. If b and c are bothpositive constants, which of the followingexpressions best describes the acceleration (A) ofthe particle ?A) a = +bB) a = -cC) a = b + ctD) a = b-ct

8. Two particles instantaneously at A and B are 5m, apart and

they are moving with uniform velocities, the former towards Bat 4m/s and the latter perpendicular to AB at 3m/s. They arenearest at the instantA) 2/5sB) 3/5sC) 1sD) 4/5s

9. Three particles start from the origin at the same time, one with a velocity 'u1',alone X-axis, the second along the Y-axis with a velocity u2 and the third alognx = y line. The velocity of third particle so that the three may always lie on thesame plane is

A)2

uu 21 + B) 21uu C)21

21uu

uu+

D)21

21

uuuu2

+10. A ball is shot vertically upwards from the surface of a planet in a distant solar

system. A plot of the y versus t for the ball isshown in fig. The magnitude of the free fall inm/s2 on the planet isA) 4B) 8C) 12D) 16

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 0406460665711. The horizontal range of a projectile is R and the maximum height attained by

it is H. A strong wind now begins to blow in the direction of the motion of theprojectile, giving it a constant horizontal acceleration = g/2. Under the sameconditions of projection, the horizontal range of the projectile will now be

A) R +2

HB) R + H C) R +

2

H3D) R + 2H

12. A particle moves int he xy plane with a constant acceleration g in the negative

y-direction. Its equation of motion is y = ax - bx2, where a and b areconstants. Which of the following are correct ?A) The x-component of its velocity is constant

B) at the origin, the y-component of its velocity is ab2

g

C) At the origin, its velocity makes an angle tan-1(a) with the x-axis.D) the particle moves exactly like a projectile.

13. Two bodies are projected simultaneously from the same point, in the samevertical plane, one towards east and other towards west with velocities 8 ms-1

and 2 ms-1 respectively. The time at which their velocities are perpendicular toeach other isA) 2/5 s B) 5/2s C) 1/5 s D) 5 s

14. Two stones are projected so as to reach the same distance from the point ofprojection on a horizontal surface. The maximum height reached by oneexceeds the other by an amount equal to half the sum of the heights attainedby them. Then the angles of projection for the stones areA) 450, 1350 B) 00, 900 C) 300, 600 D) 200, 700

15. Velocity and acceleration of a particle at some instant of time are ( )j4i3v += m/s and ( )j8i6a += m/s2 respectively. At the same instant particle is atorigin. Maximum x-co-ordinate of particle will beA) 1.5m B) 0.75m C) 2.25m D) 4.0m

16. a-t graph for a particle moving in a straight line is asshown in figure. Change in velocity of the particle fromt=0 to t=6s is :A) 10m/s B) 4m/sC) 12m/s D) 8m/s

17. Speed time graph of two cars A and B approachingtowards each other is shown in figure. Initial distancebetween them is 60m. The two cars will cross each otherafter time.

A) 2sec B) 3secC) 1.5sec D) 2 sec

18. The position of a particle along x-axis at time t is given by x = 2 + t - 3t2. Thedisplacement and the distance travelled in the interval t = 0 to t = 1 arerespectivelyA) 2, 2 B) -2, 2.5 C) 0, 2 D) -2, 2.16

19. The acceleration time graph of a particle moving alonga straight line is as shown in figure. At what time theparticle acquires its initial velocity ?

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657A) 12sB) 5sC) 8sD) 16s

20. A graph between the square of the velocity of aparticle and the distance s moved by the particle is

shown in the figure. The acceleration of the particlein kilometre per hour square is :A) 2250B) 225C) -2250D) -225

21. A particle starts from rest and traverses a distance l with uniformacceleration, then moves uniformly over a further distance 2l and finallycomes to rest after moving a further distance 3l under uniform retardation.Assuming entire motion to be rectilinear motion the ratio of average speed

over the journey to the maximum speed on its way is :A) 1/5 B) 2/5 C) 3/5 D) 4/5

22. Two stones are thrown up simultaneously with initial speeds of u1 andu2 (u2 >u1). They hit the ground after 6s and 10s respectively. Which graph in figure

correctly represents the time variation of x = (x2 - x1), the relative positionof the second stone with respect to the first upto t = 10s ? Assume that thestones do not rebound after hitting the ground.

A) B) C) D)

23. Figure shows the position-time (x-t) graph of the motion of two boys A and Breturning from their school O to their homes P and Qrespectively. Which of the following statements is true ?A) A walks faster than BB) Both A and B reach home at the same timeC) B starts for home earlier than AD) A overtakes B on his way to home

24. A ball is projected with a velocity 20 3m/s at angle 600 to the horizontal. Thetime interval after which the velocity vector will make an angle 300 to thehorizontal is (take g = 10m/s2)A) 4sec B) 2 sec C) 1 sec D) 3 sec

25. The equation of motion of a projectile is : y = 12x -4

3x2

Given that g = 10ms-2, what is the range of the projectile ?A) 12m B) 16m C) 20m D) 24m

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 0406460665726. A projectile is thrown with an initial velocity of( ) 1msjbia + . If the range of the

projectile is twice the maximum height reached by it, then :A) a = 2b B) b=a C) b = 2a D) b = 4a

27. A particle moves along a parabolic path y = 9x2 in such a way that the x

component of velocity remains constant and has a value 1ms3

1 . The

acceleration of the particle is

A) 2msj3

1 B) 2msj3 C) 2msj3

2 D) 2msj2

28. Two projectiles are projected with the same velocity. If one is projected at anangle of 300 and the other at 600 to the horizontal. The ratio of maximumheights reached, is :A) 1 : 3 B) 2 : 1 C) 3 : 1 D) 1 : 4

29. A particle is projected from the ground with velocity u at angle withhorizontal. The horizontal range, maximum height and time of flight are R, Hand T respectively. They are given by,

R =g2

sinuH,

g

2sinu222 = and T =

g

sinu2

Now keeping u as fixed, is varied from 300 to 600. Then,A) R will first increase then decrease, H will increase and T will decreaseB) R will first increase then decrease, while H and T both will increaseC) R will decrease while H and T will increaseD) R will increase while H and T will decrease

30. Velocity and acceleration of a particle at some instant of time are

( )k2ji2v += m/s and ( )kj6ia += m/s2 . Then, the speed of the particle

is ....... at a rate of ....... m/s2A) increasing, 2 B) decreasing ,2 C) increasing, 4 D) decreasing, 4

31. x and y coordinates of a particle moving in xy plane at some instant are :

x = 2t2 and y = 2t2

3

The average velocity of particle in a time interval from t = 1 second to t = 2second is :

A) ( ) s/mj5i8 + B) ( ) s/mj9i1 2 + C) ( ) s/mj5.4i6 + D) ( ) s/mj6i1 0 +

32. A particle is projected upwards with some velocity. At what height fromground should another particle be just dropped at the same time so that bothreach the ground simultaneously. Assume that first particle reaches to amaximum height H.A) 6H B) 8H C) 4H D) 10H

33. Particle A moves with 4m/s along positive y-axis and particle B in a circle x2 +

y2 = 4(anticlockwise) with constant angular velocity = 2 rad/s. At time t = 0particle is at (2m, 0). Then :A) magnitude of relative velocity between them at time t is 8 sin 2t

B) magnitude of relative velocity between them is maximum at t = /4second.

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657C) both A and B are correctD) both A and B are wrong

34. An armored car 2 m long and 3 m wide is moving at 10 ms-1 when a bullet hitsit in a direction making an angle tan-1(3/4) which the length of the car as seenby a stationary observer. The bullet enters one edge of the car at the cornerand passes out at the diagonally opposite corner. Neglecting any interaction

between the car and the bullet, the time for the bullet to cross the car isA) 0.20 s B) 0.15 s C) 0.10 s D) 0.50 s

35. The V - t graph for the rectilinear motion of a particle is represented by aparabola as shown in fig. Find the distance traveled by the particle in time T/2.

A) 3

TV2

max

B) 2

TV2

max

C) 3

TVmax

D) 2

TV2

max

36. A glass wind screen whose inclination with the vertical can be changed is

mounted on a car. The car moves horizontally with a speed of 2 m/s. At whatangle with the vertical should the wind screen be placed so that the raindrops falling vertically downwards with velocity 6 m/s strike the wind screenperpendicularly.A) tan-1 (1/3) B) tan-1 (3) C) cos-1(3) D) sin-1(1/3)

37. Two stones are thrown up simultaneously from the edge of a cliff with initialspeeds v and 2 v. The relative position of the second stone with respect to firstvaries with time till both the stones strike the ground asA) Linearly B) First linearly then parabolicallyC) Parabolically D) First parabolically then linearly

38. There are two values of time for which a projectile is at the same height. The

sum of these two times is equal toA) 3T/2 B) 4T/3 C) 3T/4 D) T

39. A particle is projected from a horizontal plane with 8 2 m/s at an angle. Athighest point its velocity is found to be 8 m/s. Its range will beA) 6.4 m B) 3.2 m C) 5 m D) 12.8 m

1 2 3 4 5 6 7 8 910

11

12

13

14

15

16

17

18

19

20

A B D D A B D D D B D A A C B B B D C C21

22

23

24

25

26

27

28

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30

31

32

33

34

35

36

37

38

39

C A B B B C D A B B C C D A C D

LEVEL - III1. Two particles A and B are connected by a rigid rod AB. The

rod slides on perpendicular rails as shown in fig. Thevelocity of A to the left is 10m/s. What is velocity of B when

= 600?A) 10m/sB) 5.8m/sC) 17.3m/s D) 9.8m/s

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 040646066572. In the figure , the pulley P moves to the right with a

constant speed u. The downward speed of A is vA, andthe speed of B to the right is vB.A) vB = vAB) vB = u + vAC) vB + u = vAD) The two blocks have accelerations of the same magnitude

3. A marble starts falling from rest on a smooth inclined plane forming an angle with horizontal. After covering distance 'h' the ball rebound off the plane.The distance from the impact point where the ball rebounds for second time is

A) 8h cos B) 8h sin C) 2h tan D) 4h sin4. From the top of a tower of height 40m, a ball is projected upwards with a

speed of 20m/s at an angle of elevation of 300. The ratio of the total timetaken by the ball to hit the ground to its time of flight (time taken to comeback to the same elevation) is (take g = 10m/s2)A) 2 : 1 B)3 : 1 C) 3 : 2 D) 1.5 : 1

5. If time taken by the projectile to reach Q is T, then PQ is equal to :

A) Tv sin

B) Tv cos

C) Tv sec

D) Tv tan

6. A particle is thrown with a speed u at an angle with the horizontal. When

the particle makes an angle with the horizontal, its speed changes to v :

A)v = u cos B) v = u cos cos C) v = u cos sec

D) v = u sec cos

7. A stone is projected from a point on the ground so as to hit a bird on the topof a vertical pole of height h and then attain a maximum height 2h above theground. If at the instant of projection the bird flies away horizontally with auniform speed and if the stone hits the bird while descending then the ratio ofthe speed of the bird to the horizontal speed of the stone is :

A)12

2

+B)

12

2

C)

2

1

2

1+ D)

12

2

+

8. Shots are fired simultaneously from the top and bottom of a

vertical cliff with the elevation = 300, = 600 respectively

and strike the object simultaneously at the same point.If a = 30 3m is the horizontal distance of the object from thecliff, then the height of the cliff is :A) 30m B) 45mC) 60m D) 90m

9. A particle if projected up an inclined plane of length 20m and inclination 300

(with horizontal). What should be the value of angle (with horizontal) withwhich the projectile be projected so that it strikes the plane exactly at mid-point, horizontally :

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A) = tan-1

3

2

B) = tan-1 (2) C) = tan-1 (3

)D) = tan-1

2

3

10. A stone is projected at an angle with the horizontal with velocity u. Itexecutes a nearly circular motion near its maximum height for a short time.The radius of circular path is :

A) g2

cosu22

B) g

sinu 22

C) g

cosu22

D) g

u2

11. Two projectiles A and B are fired simultaneously asshown in figure. They collide in air at point P at time t.then :

A) t(u1 cos 1 - u2 cos 2) = 20

B) t(u1 sin 1 - u2 sin 2) = 10C) both (A) and (B) are correctD) both (A) and (B) are wrong

12. Two stones are projected simultaneously with equal speeds from point on aninclined plane along the line of its greatest slope upwards and downwardsrepectively. The maximum distance between their points of striking the planeis double that of when they are projected on a horizontal ground with samespeed. If one strikes the plane after two seconds of the other, the angle ofinclination of plane isA) 300 B) 450 C) 350 D) 150

13. A particle is projected under gravity with velocity ga2 from a point at a

height h above the level plane. The maximum range R on the ground is

A) h)1a( 2+

B) ha2

C) ha

D) 2 )ha(a +

14. The co-ordinates of a particle moving in a plane aregiven by x = a cos pt and y = b sin pt where a, b (< a)and p are positive constants of appropriate dimensions. ThenA) The path of the particle is an ellipseB) The velocity and acceleration of the particle are normal to each other at t =

/2pC) The acceleration of the particle is always directed towards a fixed point

D) The distance traveled by the particle in time interval t = 0 to t = /2p is a.

1 2 3 4 5 6 7 8 9 10 11 12 13 14B BD B A D C D C A C B B D

MULTIPLE ANSWER TYPE QUESTIONS

1. A particle of mass m is thrown up vertically with velocity u. As air exerts aconstant force F, the particle returns back at the point of projection with velocity

vafter attaining maximum height h, then

V=

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A))m/Fg(2

uh

2

+= B)

)m/Fg(2h

2

=

vC)

)m/Fg(

)m/Fg(u

+

=v D))m/Fg(

)m/Fg(u

+

=v

2. A particle of mass m moves on X-axis as follows; it starts from rest at t = 0from the point x = 0, and comes to rest at t = 1 at the point x = 1. No otherinformation is available about its motion at intermediate times (0 < t < 1). If

denotes the instantaneous acceleration of the particle, then

A) cannot remain positive for all t in the interval 0 t 1B) | | cannot exceed 2 at any point in its path

C) | | must be 4 at some point or points in its path

D) must change sign during the motion, but no other assertion can bemade with the information given

3. Two trains are travelling along a straight track one behind the other. The firsttrain is travelling at 12 m/s. The second train, approaching from the rear istravelling at speed v> 12 m/s when the second train is 200 m behind thefirst, the driver of second train applies brakes producing a uniformdeceleration of 0.20 m/s2. ThenA) If v = 20 m/s, the trains will not collideB) If v = 20 /s, the trains will collide after about 20 sC) If v = 27 m/s, the trains will not collideD) If v = 27 m/s, the trains will collide after about 15s

4. A particle of mass m and charge q starts from rest from origin along X-axis ina region where an electric field E = E0 - a x exists. Here E0 and a are constantand x is the distance from the starting point. Then in the region between x =0 to x = 2 E0/a.A) the speed of particle first increases, then decreasesB) the particle comes to rest at x = 2 E0/aC) the particle has maximum speed at x = E0/a

D) the particle is subjected to an acceleration which changes sign at x = E0/a5. Two particles are projected from the same point with the same speed v0 at the

different angles 1 and 2 with the horizontal. Their respective times offlights are T1 and T2. If they have the same horizontal range, and theirmaximum heights are H1 and H2 respectively, then

A) 1 + 2 = 900 B) H1 + H2 = v02/2g C) 12

1 tanT

T= D)

12

2

1 tanH

H=

6. A projectile thrown on a level surface attains a height h after t1 seconds andagain after t2 seconds. If the maximum height attained by the projectile is Hafter t seconds, then

A) t1t2 = gh2

B) t1 + t2 =g

H8C) t2 - t1 =

g

)hH(8 D) t2 - t = t - t1

7. Which of the following statements are true for a moving body?A) If its speed changes, its velocity must change and it must have someaccelerationB) If its velocity changes, its speed must change and it must have someacceleration

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INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657C) If its velocity changes, its speed may or may not change, and it must havesome accelerationD) If its speed changes but direction of motion does not change, its velocitymay remain constant

8. The figure shows the velocity ( ) of a particle plottedagainst time (t)A) The particle changes its direction of motion at somepointB) The acceleration of the particle remains constantC) The displacement of the particle is zeroD) The initial and final speeds of the particle are the same

9. A particle starts from the origin of coordinates at time t = 0 and moves in the

xy plane with a constant acceleration in the y-direction. Its equation of

motion is y = x2. Its velocity component in the x-direction is

A) variable B)2

C)2

D)

2

10. Two particles A and B start simultaneously from the same point and move in ahorizontal plane. A has an initial velocity u1 due east and acceleration a1 duenorth. B has an initial velocity u2 due north and acceleration a2 due eastA) Their paths must intersect at some pointB) They must collide at some pointC) They will collide only if a1u1 = a2u2D) If u1 > u2 and a1 < a2, the particles will have the same speed at some pointof time

11. Two particles are projected from the same point with the same speed, at

different angels 1 and 2 to the horizontal. They have the same horizontalrange. Their times of flight are t1 and t2 respectively

A) 1 + 2 = 900 B) 12

1 tant

t = C) 22

1 tant

t = D)2

2

1

1

sin

t

sin

t

=

12. A large rectangular box falls vertically with an acceleration a. Atoy gun fixed at A and aimed towards C fires a particle P.A) P will hit C if a = gB) P will hit the roof BC if a > gC) P will hit the wall CD or the floor AD if a < gD) May be either A, B, or C depending on the speed of projectionof P

13. Two shells are filed from a cannon with speed u each, at angles of and respectively with the horizontal. The time interval between the shots is T. They

collide in mid air after time t from the first shot. Which of the followingconditions must be satisfied?

A) > B) t cos = (t - T) cos C) (t - T) cos = t cos

D) (usin ) t -2

1gt2 = (u sin ) (t - T) -

2

1g (t - T)2

14. A particle moving with a speed v changes direction by an angle , withoutchange in speedA) The change in the magnitude of its velocity is zero

B) The change in the magnitude of its velocity is 2v sin /2

C) The magnitude of the change in its velocity is 2 v sin /2

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D) The magnitude of the change is its velocity is v (1 - cos )15. A particle of mass 'm' moves on the x-axis as follows:

It starts from rest at t = 0 from the point x = 0 and comes to rest at t = 1 atthe point x = 1.No other information is available about its motion at intermediate times ( 0 C) (T - t) cos = t cos

D) ( u sin ) T -2

1gT2 = ( u sin ) (T - t) -

2

1g(T - t)2

33. Two guns situated at the top of a hill of height 10 m, fire one shot each withthe same speed of 5 3 ms-1 at some interval of time. One gun fireshorizontally and other fires upwards at an angle of 600 with the horizontal. Theshots collide in mid air at the point P. Taking the origin of the coordinatesystem at the foot of the hill right below the muzzle, trajectories in x - y planeand g = 10 ms-2 thenA) the first shell reaches the point P at t1 = 1 s from the start

B) the second shell reaches the point P at t2 = 2 s from the startC) the first shell is fired 1 s after the firing of the second shellD) they collide at P whose coordinates are given by (5 3 , 5) m

34. A radar observer on the ground is watching an approaching projectile. At acertain instant he has the following information:i) The projectile has reached the maximum altitude and is moving with ahorizontal velocity v;ii) The straight line distance of the observer to the projectile is l;

iii) The line of sight to the projectile is at an angle above the horizontalAssuming earth to be flat and the observer lying in the plane of theprojectile's trajectory then,A) the distance between the observer and the point of impact of the projectileis

D =g

sinv2 - l cos

B) the distance between the observer and the point of impact of the projectile is

D = vg

sin2 l- l cos

C) the projectile will pass over the observer's head for l

30432279 Jr M1 Kinematics

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