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The language of kinematics and motion graphsAQA A-Level Maths: Revision notes

Section 1

Position, displacement and distance

Position is where an object is relative to a fixed origin OO on a line, and has a sign for direction. Displacement is the change in position: a vector, with size and direction. Distance travelled is the total length of the path: a scalar, never negative. Example: a runner goes 4040 m forward then 1515 m back. Her displacement is 2525 m in the forward direction but the distance travelled is 5555 m. When the object never reverses, the distance equals the size of the displacement.

Key termspositiondisplacementdistance travelledvectorscalar
Common mistake

Using total distance when asked for displacement, or the other way round, when the object changes direction.

Section 2

Velocity, speed and acceleration

  • Velocity is the rate of change of displacement. Average velocity =displacementtime=\frac{\text{displacement}}{\text{time}}, a vector.
  • Speed is the rate of change of distance. Average speed =distancetime=\frac{\text{distance}}{\text{time}}, a scalar (the size of velocity at an instant).
  • Acceleration is the rate of change of velocity: change in velocitytime\frac{\text{change in velocity}}{\text{time}}. A negative acceleration means the velocity is decreasing (a deceleration if the object is moving in the positive direction). For the runner above, average speed over 1313 s is 5513=4.23\frac{55}{13}=4.23 m s−1^{-1} but average velocity is 2513=1.92\frac{25}{13}=1.92 m s−1^{-1}.
Key termsvelocityspeedacceleration
Exam tip

Ask yourself whether the question needs direction. If it does, use displacement and velocity; if not, use distance and speed.

Section 3

Displacement-time graphs

On a displacement-time graph:

  • the gradient is the velocity
  • a horizontal line means the object is at rest
  • a negative gradient means it is moving in the negative direction
  • a steeper line means a greater speed. Example: displacement rising from 00 to 100100 m in 2020 s has velocity 55 m s−1^{-1}; falling from 100100 m to 4040 m in 3030 s has velocity −2-2 m s−1^{-1}. A curve means the velocity is changing.
Key termsgradient
Common mistake

Reading the height of a displacement-time graph as the velocity. The height is displacement; the gradient is velocity.

Section 4

Velocity-time graphs

On a velocity-time graph:

  • the gradient is the acceleration
  • the area under the graph is the displacement (area below the axis is negative displacement)
  • a horizontal line is constant velocity, a sloping straight line is constant acceleration. Example: a train accelerates 00 to 2020 m s−1^{-1} in 1010 s, runs at 2020 m s−1^{-1} for 3030 s and stops in 2020 s. Acceleration at first is 22 m s−2^{-2}, and the distance is 12(10)(20)+30(20)+12(20)(20)=900\frac12(10)(20)+30(20)+\frac12(20)(20)=900 m.
Key termsarea under the graph
Exam tip

Split the area into triangles and rectangles, or use the trapezium formula 12(a+b)h\frac12(a+b)h with the parallel sides aa and bb the two time lengths.

Section 5

Choosing and interpreting

Link the quantity asked for to the graph feature: velocity from the gradient of a displacement-time graph, acceleration from the gradient of a velocity-time graph, displacement from the area under a velocity-time graph. For a round trip the total displacement can be zero while the distance is not, so average velocity is zero but average speed is not. State units in every answer and give direction (or a sign) for vectors.

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Exam questions on The language of kinematics and motion graphs

  1. A runner starts at point OO on a straight track and runs 4040 m in the positive direction in 88 s. She then immediately turns round and runs 1515 m back towards OO in 55 s.
    Find the average speed of the runner over the whole 1313 s, and explain why it is different from the magnitude of her average velocity.2 marks
  2. A train travels along a straight track. Its velocity-time graph is made of three straight-line sections: the velocity increases uniformly from 00 to 2020 m s−1^{-1} in 1010 s, then stays constant at 2020 m s−1^{-1} for 3030 s, then decreases uniformly to 00 in 2020 s.
    Find the acceleration of the train in the last 2020 s, and state what the sign of your answer tells you.2 marks
  3. A cyclist travels along a straight road. Her displacement ss metres from a point OO, at time tt seconds, has a displacement-time graph made of three straight-line sections: ss increases uniformly from 00 to 100100 when tt goes from 00 to 2020; ss stays at 100100 from t=20t=20 to t=50t=50; and ss decreases uniformly from 100100 to 4040 from t=50t=50 to t=80t=80.
    Find the velocity of the cyclist in each of the three stages of her journey.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).