The gradient of a velocity-time graph gives the acceleration
Motion & Kinematics - OCR A-Level Physics
- The gradientThe rate of change of one variable with respect to another; the slope of a graph at a point. of a velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.-time graph gives the 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.
- The area under the curve gives the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m).
- A straight line with positive gradient indicates 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 indicates constant velocityUnchanging speed in a fixed direction — zero acceleration. (zero accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻².)
- Area below the time axis represents displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). in the negative direction
- For non-uniform acceleration, estimate the area by counting squares or using the trapezium rule
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