Movement & Position Notes

Q: When do we use v² = u² + 2as?

Use v^2 = u^2 + 2as when you know the acceleration and distance but not the time. The equation links final speed v, initial speed u, acceleration a and distance s. It is useful in braking and stopping-distance problems.

Edexcel IGCSE Physics: Revision notes

Key facts

On a distance-time graph, the gradient is the speed; a flat line means stationary.
Average speed = distance ÷ time.
Acceleration $a = \dfrac{v - u}{t}$ , in m/s².
On a velocity-time graph, the gradient is acceleration and the area under the line is distance.
When time is not given, use $v^2 = u^2 + 2as$ .

Distance-time graphs

The gradient of a distance-time graph is the speed: flat means stationary, straight means constant speed, curved means changing speed.

A distance-time graph plots distance travelled against time. The steeper the line, the faster the object.

Average speed is the total distance divided by the total time taken.

Stationary (flat), constant speed (straight) and accelerating (curved) on a distance-time graph.

Worked example

An object moves 150 m in 20 s at a steady speed. What is its average speed?

1
Average speed $=$ distance ÷ time.
2
$150 \div 20$ .

What does a horizontal line on a distance-time graph show?

Investigating motion

Measure distance at regular time intervals, repeat and average, and use light gates to avoid reaction-time error.

Toy cars or trolleys can be timed over measured distances. Repeating readings and averaging improves reliability.

A stopwatch adds human reaction-time error; light gates start and stop the timer automatically and are more accurate.

1
Measure
distance at regular time intervals
2
Time it
stopwatch, or light gates for accuracy
3
Repeat
average to improve reliability
4
Plot
a distance-time or velocity-time graph

Investigating motion

Why are light gates better than a stopwatch?

Velocity-time graphs

On a velocity-time graph, the gradient is the acceleration and the area under the line is the distance travelled.

Acceleration is the rate of change of velocity, $a = \frac{v - u}{t}$ , in m/s². Here $u$ is initial and $v$ is final velocity.

On a velocity-time graph, flat means constant velocity, upwards means speeding up and downwards means slowing down.

A car accelerating from rest: the shaded area is the distance travelled.

Worked example

A car accelerates from 0 to 20 m/s in 5 s. Find its acceleration and distance.

1
$a = \frac{20 - 0}{5} = 4$ m/s².
2
Area under the line: $\frac{1}{2} \times 5 \times 20$ .

u = 0start velocity ua = 2acceleration a

Change u and a to see how the line changes.

A velocity-time graph is a horizontal line at 8 m/s. What is the acceleration?

The equation v² = u² + 2as

Use $v^2 = u^2 + 2as$ when you know the acceleration and distance but not the time.

The equation links final speed $v$ , initial speed $u$ , acceleration $a$ and distance $s$ . It is useful in braking and stopping-distance problems.

This equation of motion is only for constant acceleration.

u = 4initial speed u (m/s)a = 2acceleration a (m/s²)

v² = u² + 2as plotted as v² against s (illustrative values): the intercept is u² and the gradient is 2a

Equation of motion $v^2 = u^2 + 2as$

Worked example

A car at $u = 10$ m/s accelerates at 2 m/s² over 24 m. Find its final speed.

1
$v^2 = 10^2 + 2 \times 2 \times 24 = 196$ .
2
$v = \sqrt{196}$ .

A ball starts from rest and accelerates at 5 m/s² over 10 m. What is its final speed?

Try an exam question

A car accelerates uniformly from rest to 20 m/s in 5 s. (a) Calculate its acceleration. (b) Calculate the distance travelled in this time.

[4 marks]

That's the notes covered.

Carry on to the next subtopic.