A particle moves from A to B under constant acceleration.
The distance between A and B is sss.
The speed at A is uuu. The speed at B is vvv.
The time taken is ttt.
The acceleration is aaa.
Given s=72 ms = 72 \text{ m}s=72 m, u=0 ms−1u = 0 \text{ ms}^{-1}u=0 ms−1, v=24 ms−1v = 24 \text{ ms}^{-1}v=24 ms−1. Find a a\,a and ttt.
Given s=144 ms = 144 \text{ m}s=144 m, a=2 ms−2a = 2 \text{ ms}^{-2}a=2 ms−2, v=24 ms−1v = 24 \text{ ms}^{-1}v=24 ms−1. Find t t\,t and uuu.
Given s=56 ms = 56 \text{ m}s=56 m, t=4 st = 4 \text{ s}t=4 s, v=20 ms−1v = 20 \text{ ms}^{-1}v=20 ms−1. Find a a\,a and uuu.
Given s=84 ms = 84 \text{ m}s=84 m, a=2 ms−2a = 2 \text{ ms}^{-2}a=2 ms−2, t=4 st = 4 \text{ s}t=4 s. Find u u\,u and vvv.
Given v=14 ms−1v = 14 \text{ ms}^{-1}v=14 ms−1, a=2 ms−2a = 2 \text{ ms}^{-2}a=2 ms−2, t=3 st = 3 \text{ s}t=3 s. Find s s\,s and uuu.
197 exam-style questions on Edexcel A Level Maths Constant Acceleration, covering 8.2 Modelling Assumptions, 8.3 Quantities and Units, 8.4 Working with Vectors, 9.1 Displacement-Time Graphs, 9.2 Velocity-Time Graphs, 9.3 Constant Acceleration Formulae 1, 9.4 Constant Acceleration Formulae 2, and 9.5 Vertical Motion Under Gravity. Each one has a worked solution and a mark scheme showing where the marks go.