The gradient of a velocity-time graph gives the acceleration
Motion & Kinematics - OCR A-Level Physics
Two rules in one graph
The gradient of a velocity-time graph gives the accelerationThe rate of change of velocity. A vector. Measured in m s⁻².. The area under the graph gives the displacementThe distance moved in a particular direction from a starting point. A vector. Measured in m.. These two readings are the most heavily examined skill in motion.
$$a = \frac{\Delta v}{\Delta t}, \qquad s = \text{area under } v\text{-}t \text{ graph}$$
Diagram pending
Velocity-time graph showing three regions: rising straight line (uniform acceleration, triangular area shaded), horizontal line (constant velocity, rectangular area shaded), falling straight line (deceleration). Labels mark the gradient triangle on each segment and the displacement = area under graph.
Will be replaced with a GeoGebra SVG in stream 2.
Case 1: Straight line, positive gradient
Uniform acceleration. Slope = $a$ in $\text{m s}^{-2}$. Area under it is a triangle (if it starts at $v = 0$) or a trapezium: $s = \frac{(u + v)}{2} \times t$.
Case 2: Horizontal line
Constant velocityUnchanging speed in a fixed direction; zero acceleration., zero acceleration. Area is a rectangle: $s = v \times t$.
Case 3: Negative gradient
Deceleration. $a$ is negative. The line slopes downward. When it reaches $v = 0$, the object is instantaneously at rest. If the line crosses below the axis, the object has reversed direction.
Case 4: Curve
Non-uniform acceleration. For instantaneous $a$, draw a tangent and find its gradient. For displacement, estimate the area by counting squares or applying the trapezium rule.
- Area above the time axis is positive displacement; area below is negative displacement. For total displacement, subtract. For total distance, add the magnitudes.
- Trapezium rule for one segment: $\text{area} = \tfrac{1}{2}(v_1 + v_2) \times \Delta t$. Sum the trapezia for the full graph.
- A rocket accelerating from $0$ to $50 \text{ m s}^{-1}$ in $10 \text{ s}$ has $a = 5 \text{ m s}^{-2}$ and travels $\tfrac{1}{2} \times 10 \times 50 = 250 \text{ m}$.
Common Mistake
MEDIUM
Students often: It's easy to forget the area under a v-t graph below the time axis counts as negative displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m)..
Instead: When calculating total displacement, subtract the area below the axis. For total distance, add the magnitudes of all areas.
Instead: When calculating total displacement, subtract the area below the axis. For total distance, add the magnitudes of all areas.
Worked Example [3 marks]
A car accelerates uniformly from rest to 20 $m s^{-1}$ in 8.0 s, then travels at constant velocity for 12 s. Calculate the total displacement.
Show Solution
1
$Displacement during acceleration = area of triangle = \frac{1}{2} \times 8.0 \times 20 = 80 \text{ m}$
[1]2
$Displacement at constant velocity = area of rectangle = 20 \times 12 = 240 \text{ m}$
[1]3
$Total displacement = 80 + 240 = 320 \text{ m}$
[1]Answer
320 m