Motion with constant acceleration
Choose a kinematics equation and test its assumptions.
Before you begin
- Velocity and signed quantities
- Rearranging formulae
Define a direction and an interval
Acceleration is the rate of change of velocity. Choose a positive direction and use signed velocities consistently. A negative acceleration may mean slowing down or speeding up, depending on velocity's direction. The familiar constant-acceleration equations apply only while acceleration is constant over the chosen interval.
Connect equations to a velocity-time graph
For constant acceleration, velocity changes linearly with time, so the mean velocity is (u + v)/2. Displacement is the signed area under the velocity-time graph. Combining this with v = u + at gives s = ut + ½at². Displacement can differ from distance if direction reverses during the interval.
Worked example
A trolley starts at 2 m/s and accelerates at 3 m/s² for 4 s in the same direction. Find its final velocity and displacement.
Show the worked solution
- Final velocity is 2 + 3 × 4 = 14 m/s.
- Displacement is 2 × 4 + ½ × 3 × 4².
- Add 8 + 24 = 32 m and check using mean velocity 8 m/s for 4 s.
14 m/s; displacement 32 m.
Try it yourself
A velocity-time graph has constant positive slope. What does that indicate?
Common mistakes
- Do not use constant-acceleration equations through an interval with varying acceleration.
What can you explain now?
A car slows uniformly from 20 m/s to rest in 5 s. Find its acceleration and stopping displacement.
Compare with the explanation
−4 m/s² and 50 m.
Taking the initial direction as positive, a = (0 − 20)/5. Mean velocity is (20 + 0)/2 = 10 m/s, giving 10 × 5 = 50 m.
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