3.4.1.3
Velocity-time graphs: $gradient = acceleration, area = displacement$
Motion Along a Straight Line — AQA A-Level Physics
- A straight $line = uniform (constant) accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².$.
- A horizontal $line = constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. (zero accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².)$.
- A $curve = non-uniform accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻².$.
- Positive $gradient = acceleration. Negative gradient = deceleration$.
- The area under the curve equals the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (or distance if speed-time).
- Split the area into triangles, rectangles and trapeziums to calculate displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m)..
Worked Example
A vehicle accelerates uniformly from rest. At 40 s its velocity is 105 km h^-1. Calculate the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). at 40 s.
Show Solution
1
Convert velocity to km s^-1
$$105 \text{ km h}^{-1} = \frac{105}{3600} = 0.0292 \text{ km s}^{-1}$$
2
Calculate area under the graph (triangle)
$$s = \frac{1}{2} \times 40 \times 0.0292 = 0.58 \text{ km} \approx 600 \text{ m}$$
Answer
Displacement is approximately 600 m.
Examiner Tips and Tricks
- Always check the y-axis units.
- Students frequently confuse displacement-time and velocity-time graphs.
- The area under a v-t graph gives displacement; the area under an a-t graph gives change in velocity.