1

Using Kinematic Equations

1. Write down the symbols, values and units (in SI) of given quantities

2. Write down the symbol of the quantities required

3. Select the equation that contains all of the symbols in 1. and 2. above

e.g.

A stone is released from a height of 20.0 m above the ground. Neglecting air resistance and using the acceleration due to gravity as 9.81 ms-2, find the velocity with which the stone will hit the ground .

vf2 = vi

2 + 2ad

vf2 = 02 + 2 x 9.81 x 20

vf = 392 = 19.8 m s-1

vi = 0 from rest

d = 20.0 m

a = 9.81 ms-2

vf = ?vf

2 = 392

©John Parkinson

2

3

Distance travelled - d

Time taken - tVelocity - v

v =d

t d

v | t

Velocity = Speed in a Specified Direction

Constant Velocity

4

N

100 m

in 4 seconds

Distance travelled = ? 100 m

Displacement = ? 100 m to the East

Speed = ? Speed = 100/4 = 25 m s-1

Velocity = ? Velocity = 25 m s-1 to the East

5

Displacement / time graphs

Constant velocity

Displacement d

Time t

What will the graph look like?

Gradient = ?

Δt

Δd

t

dv

Velocity

6

Displacement - d

Time - t

What about this graph?

A body at rest

Displacement d

Time t

And this graph?

The gradient is …….?increasing

Δd

Δt

The body must be ……..?accelerating

7

1 3

2

A

Velocity/time graphs

Velocity - v

Time - t

Velocity - v

Time - t

This body has a constant or uniform ………?acceleration

Δv

Δt

The gradient = ?the acceleration

t

va

1 = …… ?Uniform acceleration

2 = …… ?Constant velocity

3 = …… ?Uniform retardation [deceleration]

Area under the graph = A

= …….. ?Distance travelled

8

Velocity v (ms-1)

Time t(s)

30

20 50 80

Question The graph represents the motion of a tube train between two stations

Find

1. The acceleration

2. The maximum velocity

3. The retardation

4. The distance travelled

1. The acceleration = the initial gradient = 30÷20 = 1.5 m s-2

2. The maximum velocity is read from the graph = 30 m s-1

3. The retardation = the final gradient = -30 ÷ [80-50] = -1.0 m s-2

4. The distance travelled = the area under the graph

=½ x 20 x 30 + 30 x 30 + ½ x 30 x 30 = 1650 m

9

Velocity

Time

What might this graph represent?

Can you draw an acceleration time graph for this motion?

Terminal Velocity

Acceleration

Time

9.81 m s-1

Find the acceleration at each partFind the acceleration at each part

Time (s)01234

3 6

9 12 15

5

-1-2-3-4

5 18

17

Velocity (ms-1)0-3 s: a =1 ms-2, accelerating3-5 s: a = 0 ms-2, constant velocity

6-9 s: a = 0 ms-2, v=0, rest5-6 s: a = -3 ms-2 decelerating

9-12 s: a = - 1.3 ms-2 accelerating to –ve direction12-15 s: a = 0 ms-2 , constant velocity, -ve direction

15-17 s: a = 2 ms-2 , decelerating

Find out the displacementFind out the displacement

0-6 s: d1 =(6+2)x3/2=12m, to +ve direction

Time (s)01234

3 6

9 12 15

5

-1-2-3-4

5 18

17+

–

Velocity (ms-1)

6-9 s: rest, d2 = 09-17 s: d3 =- (8+3)x4/2=-22m, to -ve direction

Total displacement d = d1 + d2 + d3= -10m, to -ve directionTotal distance =

34mAverage speed = 34/17 = 2ms-1

Average velocity = -10/17 = -0.59 ms-1