Describing Motion Notes

Edexcel GCSE Physics: Revision notes

Key facts

A scalar has size only (distance, speed); a vector has size and direction (displacement, velocity).
Speed = distance ÷ time. On a distance-time graph, gradient = speed.
Acceleration $a=\dfrac{v-u}{t}$ in m/s²; $v^2-u^2=2ax$ when time is not given.
On a velocity-time graph, gradient = acceleration and area = distance.
Light gates measure speed; $g=10$ m/s² (N/kg) in free fall.

Scalars and vectors

A scalar has only a size; a vector has a size and a direction.

Scalars: distance, speed, mass, energy.

Vectors: displacement, velocity, acceleration, force, weight, momentum.

Velocity is speed in a stated direction.

Walk 3 m east then 4 m north: the distance is 7 m, but the displacement is 5 m in the direction shown.

Walking 3 m east then 4 m north: distance 7 m, displacement 5 m.

Scalar Vector Has Size only Size and direction Examples Distance, speed, mass, energy Displacement, velocity, force, momentum

	Scalar	Vector
Has	Size only	Size and direction
Examples	Distance, speed, mass, energy	Displacement, velocity, force, momentum

Scalar

Has:: Size only
Examples:: Distance, speed, mass, energy

Vector

Has:: Size and direction
Examples:: Displacement, velocity, force, momentum

Which of these is a vector?

Speed and distance-time graphs

Average speed is distance divided by time, and it is the gradient of a distance-time graph.

Use metres and seconds: speed in m/s.

On a distance-time graph:

gradient = speed
steeper means faster
flat means stationary
a curve means the speed is changing

v = 2speed

Distance-time: the straight line is constant speed (drag v); the dashed curve gets steeper, so it is accelerating.

Average speed $\dfrac{\text{distance}}{\text{time}}$

Worked example

A runner covers 100 m in 20 s. Find the average speed.

1
Speed = 100 ÷ 20.

A runner covers 100 m in 20 s. What is the average speed?

Acceleration

Acceleration is the change in velocity divided by the time taken.

Acceleration is the rate of change of velocity, in m/s². Here $v$ is final velocity, $u$ initial velocity and $t$ time.

When the time is not given, use $v^2-u^2=2ax$ , where $x$ is the distance.

A negative acceleration is a deceleration.

Acceleration $a=\dfrac{v-u}{t}$
No time given $v^2-u^2=2ax$

Worked example

A car speeds up from 5 m/s to 25 m/s in 4 s. Find the acceleration.

1
a = (25 − 5) ÷ 4.

Worked example

A car starts from rest and accelerates at 3 m/s² over 24 m. Find its final speed.

1
v² = 0 + 2 × 3 × 24 = 144.
2
v = √144.

A cyclist slows from 10 m/s to 4 m/s in 3 s. What is the acceleration?

Velocity-time graphs

The gradient gives the acceleration and the area under the line gives the distance travelled.

On a velocity-time graph:

a horizontal line is constant velocity
a negative gradient is deceleration
the steeper line is the greater acceleration

Split the area into rectangles and triangles, or use a trapezium.

A car speeding up from 5 m/s to 25 m/s in 4 s. Gradient 5 m/s²; the shaded area is the distance.

Gradientacceleration
Area under the linedistance

Worked example

Find the distance travelled from the graph.

1
The area is a trapezium.
2
½ × (5 + 25) × 4.

A velocity-time graph is a horizontal line at 6 m/s for 5 s. How far is travelled?

Measuring speeds

A light gate times how long a card blocks the beam, and speed is card length divided by that time.

The speed through a light gate is the length of the card ÷ the time it blocks the beam.

Sound in air travels at about 330 m/s.

In free fall, the acceleration due to gravity is $g=10$ m/s². Everyday accelerations are a few m/s².

Typical everyday speeds (sound in air, 330 m/s, is left off so the others stay visible).

1
Attach a card
Of known length
2
Pass the gate
The card blocks the beam
3
Read the time
From the timer
4
Calculate
Speed = length ÷ time

Measuring speed with a light gate

A 0.05 m card blocks a light gate beam for 0.020 s. What is its speed?

Try an exam question

A car speeds up from 5 m/s to 25 m/s in 4 s. (a) Calculate the acceleration of the car. (b) Calculate the distance travelled in the 4 s, using a velocity-time graph.

[4 marks]

That's the notes covered.

Carry on to the next subtopic.