# Speed Time Graph

A Speed time graph represents the movement of an object from its initial position to its final destination

« Previous Next »## Example 1

A sprinter rulls a race of 200m. Her total time for running the race is 25 seconds ending at U m/s. Below is a sketch of the motion of the sprinter.

Calculate the

(a) acceleration in the first 4 seconds,

(b) distance covered in the first 20 seconds,

(c) value of U.

Solutions

(a)

##### Formula for Acceleration

Acceleration = $\frac{\mathrm{V\; -\; U}}{\mathrm{t}}$

V = final velocity, U = initial velocity and t = time

V = 9, U = 0 , t = 4

Acceleration = $\frac{\mathrm{9\; -\; 0}}{4}$

Acceleration = $\frac{9}{4}$

Answer: Acceleration = ${\mathrm{2.25m/s}}^{2}$

(b)

We will use a Tripezium Shape taken in the first 20 seconds to find the distance

Distance = $\frac{\mathrm{a\; +\; b}}{2}\mathrm{h}$

a = 20, b = 16, h = 9

Distance = $\frac{\mathrm{20\; +\; 16}}{2}9$

Distance = $\frac{\mathrm{36}}{2}9$

Distance = $18\times 9$

Answer: Distance = $\mathrm{162m}$

(c)

From 20 to 25 second we have another Tripezium shape, hence with the distance we have found for the first 20 second in (b) we can subtract it into the total 200m inorder to find the distance between 20 to 25 seconds

Distance = 200 - 162

Distance = 38

Distance = $\frac{\mathrm{a\; +\; b}}{2}\mathrm{h}$

Distance = 38, a = 9, b = U, h = 5

38 = $\frac{\mathrm{9\; +\; U}}{2}5$

38 = $\frac{\mathrm{45\; +\; 5U}}{2}$

Cross muiltiply

76 = $\mathrm{45\; +\; 5U}$

$\mathrm{45\; +\; 5U}$ = 76

$\mathrm{5U}$ = 76 - 45

$\mathrm{5U}$ = 31

$\frac{\mathrm{5U}}{5}=\frac{\mathrm{31}}{5}$

Answer: $U={\mathrm{6.2m/s}}^{2}$

## Example 2

The diagram below shows the speed-time graph of a 100m sprinter who accelerates uniformly for 3 seconds until he reaches a speed of 12m/s. He maintains the speed for 7 seconds and then uniformly retards for a further 4 seconds and comes to a stop.

Calculate the

(a) acceleration during the first 3 seconds,

(b) retardation at the end of his race,

(c) distance he covered in the last 4 seconds.

Solutions

(a)

Acceleration = $\frac{\mathrm{V\; -\; U}}{\mathrm{t}}$

V = 12, U = 0, t = 3

Acceleration = $\frac{\mathrm{12\; -\; 0}}{3}$

Acceleration = $\frac{\mathrm{12}}{3}$

Answer: Acceleration = ${\mathrm{4m/s}}^{2}$

You can also use the gradient formula to find the acceleration aas shown below

Gradient = $\frac{\mathrm{Y2\; -\; Y1}}{\mathrm{X2\; -\; X1}}$

We have the following co-ordinates (0, 0) (3, 12)

X1 = 0, X2 = 3, Y1 = 0, Y2 = 12

Gradient = $\frac{\mathrm{12\; -\; 0}}{\mathrm{3\; -\; 0}}$

Gradient = $\frac{\mathrm{12}}{3}$

Gradient = ${\mathrm{4m/s}}^{2}$

Therefore Answer: Acceleration = ${\mathrm{4m/s}}^{2}$

(b)

retardation is a negative acceleration

Acceleration = $\frac{\mathrm{V\; -\; U}}{\mathrm{t}}$

V = 0, U = 12, t = 4

Acceleration = $\frac{\mathrm{0\; -\; 12}}{4}$

Acceleration = $\frac{\mathrm{-\; 12}}{4}$

Acceleration = ${\mathrm{-3m/s}}^{2}$

Answer: Retardation = ${\mathrm{-3m/s}}^{2}$

(c)

We will use the area of a triangle to find the distance in the last 4 seconds

Distance = $\frac{1}{2}\mathrm{bh}$

b = 4, h = 12

Distance = $\frac{1}{2}4\times 12$

Distance = $\frac{\mathrm{48}}{2}$

Answer: Distance = $\mathrm{24m}$

## How to solve a Speed time graph

The diagram below shows the speed-time graph of a particle. the particle started off from rest and accelerated uniformly from 10 seconds. it then travelled at a constant speed of 20 seconds and then decelerated to rest.

(a) Find the speed V the particle reached if its acceleration was ${\mathrm{2m/s}}^{2}$ in the first 10 seconds.

(b) Given that the total distance covered was 750m, find the value of t in the diagram

(c) What was the speed at 40 seconds

Solution:

(a) Formula:

a = $\frac{\mathrm{V\; -\; U}}{\mathrm{t}}$## Points to note

a = acceleration

V = Final velocity or speed

U = Initial velocity or speed

t = time

Therefore a = 2, V = ? , U = 0, t = 10

2 = $\frac{\mathrm{V\; -\; 0}}{10}$

2 = $\frac{V}{10}$

Answer: V = 20m/s

(b)

The distance on the speed-time graph is given has 750m, therefore the formula should be made to fit on all the three segments of the diagram

Distance = Area1 + Area2 + Area3

Area1 = $\frac{1}{2}\mathrm{bh}$

Area1 = $\frac{1}{2}10\times 20$Area1 = $\frac{\mathrm{200}}{2}$

Area1 = $100$

Area2 = bh

Area2 = $\mathrm{20}\times 20$Area2 = $400$

Area3 = $\frac{1}{2}\mathrm{bh}$

Area3 = $\frac{1}{2}\mathrm{t}\times 20$Area3 = $\mathrm{10t}$

Distance = Area1 + Area2 + Area3

## Points to note

Distance = 750

Area1 = 100

Area2 = 400

Area3 = 10t

750 = 100 + 400 + 10t

750 = 500 + 10t

750- 500 = 10t

250 = 10t

$\frac{\mathrm{250}}{10}=\frac{t}{10}$

Answer: t = 25 s

(c)

a = $\frac{\mathrm{V\; -\; U}}{\mathrm{t}}$Note:

a = ?

V = 0

U = 20

t = 25

a = $\frac{\mathrm{0\; -\; 20}}{25}$

a = $\frac{\mathrm{-\; 20}}{25}$

a = $-0.8{\mathrm{m/s}}^{2}$

velocity = $\mathrm{acceleration}\times \mathrm{time}$

V = at

## Points to note

V = ?

a = $-0.8{\mathrm{m/s}}^{2}$ negative sign means its retardation

t = 40-25

t = 15

velocity = $a\times t$velocity = $\mathrm{0.8}\times \mathrm{15}$

Therefore the speed at 40 seconds = 12m/s