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# Kinematics - PowerPoint PPT Presentation

Kinematics. JUNIOR EAMCET. Distance and Displacement. B. 5m. 3m. A. O. 4m. O to B: distance is 7m and displacement is 5m. Distance and Displacement. B. C. 5m. 3m. A. O. 4m. O to C along OABC: distance is 11m and displacement is 3m. Distance and Displacement. B. C. 5m. 3m. A.

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### Kinematics

JUNIOR EAMCET

B

5m

3m

A

O

4m

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

B

C

5m

3m

A

O

4m

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

B

C

5m

3m

A

O

4m

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

B

A

R

R

Arc length AB is R

AB jaw length is

v

u

t

t

u

v

s

s

Both averge speed and average velocity

u

s

v

s

Both averge speed

Both averge velocity

u

Time of crossing the post is

s

Train

Post

u

Time of crossing the bridge is

L

s

Train

Bridge

Average speed =

u

B

A

Average velocity =

R

u

Average acceleration =

Acceleration is positive if v > u

a

u

v

Direction of acceleration is v – u

Acceleration is negative if v < u

u

v

a

Downward direction is positive

+ g

Freely falling body

S, u, v, g, h are all positive

Upward direction is positive

Body projecte up

a = – g

u, v, h are all positive

Upward direction is positive

u

a = – g

u is positive

s, g are negative

s = – h

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

Displacement = average velocity × time

v

u

s

If u = 0

A travels with acceleration a, B with uniform velocity u

A

u

If they start simultaneously

B

Time of meet is

u

a

d

A

If A is ahead of B

B

u

If B is ahead of A

a

d

A

B

Time of meet is

u = 0

h

Height of meet is

u = u

h = h1 + h2

Time taken to reach the ground is t1 when throuwn up

Time taken to reach the ground is t2 when thrown down

t1

u

u

t2

h

Time of free fall is

Initial velocity u =

Height h =

(n – 1)t =

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

h

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

u velocity at the top of the window

t time of crossing the window

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

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

x

v

Graphical Representation velocities u and v, the center will cross the same post with velocity ----

v

Uniform velocity

10m/s

a

-a

9

O

t

7

3

5

a

-5

Graphical Representation velocities u and v, the center will cross the same post with velocity ----

v

Uniform velocity

10m/s

-5

10/3

20

15

10

9

O

-5

t

7

3

5

a

-5

Graphical Representation velocities u and v, the center will cross the same post with velocity ----

v

vmax

O

t

V – t Graph velocities u and v, the center will cross the same post with velocity ----

The slope gives acceleration

Acceleration is positive if  < 90

v

Acceleration is negative if  > 90

P

Area represents displacement

O

Displacement is positive if area is above x - axis

t

Graphical Representation velocities u and v, the center will cross the same post with velocity ----

v

a

u constant

u

u = 0

-a

a

O

t

u velocities u and v, the center will cross the same post with velocity ----

s

A

B

Distance is 2s

Displacement is zero

v

Time for forward journey is

Average speed is

Time for return journey is

u velocities u and v, the center will cross the same post with velocity ----

t

a

u

t

a

d

Equations of Motion velocities u and v, the center will cross the same post with velocity ----

X = a + bt + ct2

Problem velocities u and v, the center will cross the same post with velocity ----

Find the acceleration

Problem velocities u and v, the center will cross the same post with velocity ----