Kinematics Kinematics

JUNIOR EAMCETJUNIOR EAMCET

Distance and DisplacementDistance and Displacement

O A

B

4m

3m5m

O to B: distance is 7m and displacement is 5m

Distance and DisplacementDistance and Displacement

O A

BC

4m

3m5m

O to C along OABC: distance is 11m and displacement is 3m

Distance and DisplacementDistance and Displacement

O A

BC

4m

3m5m

O to O along OABCO: distance is 14m and displacement is zero

Arc of a Circle Arc of a Circle

RR

A B

Arc length AB is R

AB jaw length is

Average Speed and Average Average Speed and Average Velocity Velocity

u v

t t

v

s

u

s

Both averge speed and average velocity

Average Speed and Average Average Speed and Average Velocity Velocity

vs

u

s

Both averge speed

Both averge velocity

Train Crossing a BridgeTrain Crossing a Bridge

Post Train

Bridge Train

usTime of crossing

the post is

Time of crossing the bridge is

us L

Average Speed and Averge Average Speed and Averge Velocity Velocity

ABR

Average speed =

Average velocity =

Average acceleration =

u

u

Direction of AccelerationDirection of Acceleration

u v

u v

a

a

Direction of acceleration is v – u

Acceleration is positive if v > u

Acceleration is negative if v < u

Sign Application Sign Application

+ g Downward direction is positive

Freely falling body S, u, v, g, h are all positive

Body projecte up Upward direction is positive a = – g

u, v, h are all positive

Body Projected From the Top of Body Projected From the Top of a Towera Tower

u

Upward direction is positive

a = – g s = – h

u is positive

s, g are negative

Tossing in a TrainTossing in a Train

If Vt = Vb : a = 0; the ball falls in the hand

If Vt > Vb : a > 0 the ball falls bedhind

If Vt < Vb : a < 0; the ball falls in front

Vn – Vn-1 = a Vn = u + an

Vn-1 = u + a(n – 1)

Sn – Sn-1 = a

u v

s

tDisplacement = average velocity × time

If u = 0

A

B

a

u

ABa

u

A Ba

u

A travels with acceleration a, B with uniform velocity u If they start simultaneously Time of meet is

If A is ahead of B

d

dIf B is ahead of A

One Body Projected and the One Body Projected and the Other Falling FreelyOther Falling Freely

u = 0

u = u

Time of meet is

hHeight of meet is

h = h1 + h2

u u

h

Time taken to reach the ground is t1 when throuwn up

Time taken to reach the ground is t2 when thrown down

t1

t2

Time of free fall is

Initial velocity u =

Height h =

Water Drops from the TapWater Drops from the Tap

h

(n – 1)t =

‘t’ is time interval with which the drops are released

Ratio of displacements of 4th, 3rd, 2nd and 1st drops is 1 : 3 : 5 : 7

h height of the window

u velocity at the top of the windowt time of crossing the window

If the velocity is reduced by after travelling a distnce x, then the total distance it can travel is

n

v

If th front and back of the train cross a post with velocities u and v, the center will cross the same post with velocity ----

u

v

x

Graphical RepresentationGraphical Representation

v

t

Uniform velocity

a

aO

-a10m/s

3 5 79

-5

Graphical RepresentationGraphical Representation

v

t

Uniform velocity

10/3

aO

-510m/s

3 5 79

-5

15 20 10-5

Graphical RepresentationGraphical Representation

v

tO

vmax

v

tO

P

V – t GraphV – t GraphThe slope gives acceleration

Acceleration is positive if < 90

Acceleration is negative if > 90

Area represents displacement

Displacement is positive if area is above x - axis

Graphical RepresentationGraphical Representation

v

t

u constant

au = 0

a

u

-a

O

A Bsu

vDistance is 2s

Displacement is zero

Time for forward journey is

Time for return journey is

Average speed is

u

at

u

at

d

Equations of MotionEquations of Motion

X = a + bt + ct2

Problem Problem

Find the acceleration

Problem Problem