These eight questions are original. They follow the order of the chapter, from v and a at a time to distance with turning points. Attempt each with full working before opening the answer.
Questions
Question 1. A particle has displacement s = 5t² − t³ metres. Find its velocity and acceleration when t = 2.
Answer
v = 10t − 3t² and a = 10 − 6t.
At t = 2: v = 20 − 12 = 8 m/s and a = 10 − 12 = −2 m/s².
The velocity is positive and the acceleration negative, so the particle is slowing down.
Question 2. A particle has velocity v = 6t − 9 m/s and s = 0 when t = 0. Find when it is at rest, and its displacement from O then.
Answer
At rest: 6t − 9 = 0, so t = 1.5 s.
s = 3t² − 9t. At t = 1.5: 6.75 − 13.5 = −6.75 m.
The particle is 6.75 m from O on the negative side.
Question 3. A particle has acceleration a = 6t − 4 m/s², and v = 5 m/s when t = 0. Find its velocity when t = 2.
Answer
v = 3t² − 4t + c, and v = 5 when t = 0 gives c = 5.
So v = 3t² − 4t + 5. At t = 2: 12 − 8 + 5 = 9 m/s.
Question 4. A particle has velocity v = 3t² − 18t + 24 m/s for 0 ≤ t ≤ 5. Find the displacement and the total distance.
Answer
Integrating v gives s = t³ − 9t² + 24t (taking s = 0 at t = 0). Also v = 3(t − 2)(t − 4), so it is at rest at t = 2 and t = 4.
s(0) = 0, s(2) = 8 − 36 + 48 = 20, s(4) = 64 − 144 + 96 = 16 and s(5) = 125 − 225 + 120 = 20.
Displacement = 20 − 0 = 20 m.
Distance = 20 + |16 − 20| + |20 − 16| = 20 + 4 + 4 = 28 m.
Question 5. A particle has velocity v = (t − 3)² m/s for t ≥ 0. Is the particle at rest at t = 3, and does it change direction? Find the displacement from t = 0 to t = 3.
Answer
At t = 3, v = 0, so it is at instantaneous rest.
But v ≥ 0 on both sides (for example v = 1 at t = 2 and t = 4), so the sign does not change. It does not reverse.
Displacement = ∫ (t² − 6t + 9) dt from 0 to 3 = [t³/3 − 3t² + 9t] = 9 − 27 + 27 = 9 m.
Question 6. A particle has displacement s = t² − 10t + 16 metres from O. Find (a) when it passes through O, (b) when it is at rest and its displacement then, and (c) the total distance in the first 7 seconds.
Answer
(a) s = (t − 2)(t − 8) = 0, so it passes O at t = 2 s and t = 8 s.
(b) v = 2t − 10 = 0 at t = 5 s, where s = 25 − 50 + 16 = −9 m.
(c) s(0) = 16, s(5) = −9 and s(7) = 49 − 70 + 16 = −5. Distance = |−9 − 16| + |−5 − (−9)| = 25 + 4 = 29 m.
Question 7. A particle has velocity v = 2t² − 8t m/s for t ≥ 0. Find the minimum velocity.
Answer
a = dv/dt = 4t − 8 = 0 at t = 2.
The velocity is smallest where a changes from negative to positive, because v is a parabola opening upwards.
v = 8 − 16 = −8 m/s, the minimum.
Question 8. A car brakes with velocity v = 20 − 5t m/s for 0 ≤ t ≤ 4. Find its deceleration and the distance it travels before stopping.
Answer
a = dv/dt = −5, so the deceleration is 5 m/s².
The car stops when 20 − 5t = 0, so t = 4 s.
Distance = ∫ (20 − 5t) dt from 0 to 4 = [20t − 2.5t²] = 80 − 40 = 40 m.
If you got these wrong
- Questions 1 to 3: read relating displacement, velocity and acceleration and finding displacement from a velocity function.
- Questions 4 and 6: study total distance versus displacement.
- Questions 5 and 7: work through direction changes and instantaneous rest.
- Question 8: review integration for displacement and distance.
Record each slip in the mistake log and retry after a few days. The timed practice session builder can set up a timed round.
For a teacher to watch your working, see online one-to-one Additional Mathematics tuition.