Reading a Test Trolley's Motion Model
A test trolley runs on a straight horizontal track. Its displacement s metres from a marker O at time t seconds is modelled by s = t³ − 9t² + 24t for 0 ≤ t ≤ 6, after which a separate braking control system takes over. A fixed buffer sits at s = 40 m, and any contact with it ends the test. The trolley carries an accelerometer but its data logger records only velocity, so the engineer must reconstruct the motion from the displacement model. Two claims are disputed at the debrief. A trainee reports that the trolley 'stopped at t = 3 s, because the acceleration was zero there'. The workshop foreman argues that since the trolley was at s = 16 m at t = 4 s and back at s = 36 m at t = 6 s, the total distance travelled over the six seconds is simply the change in s, namely 36 m. The engineer also needs to know when the trolley reverses and whether the buffer is reached.
1. Find expressions for velocity and acceleration, determine the times when the trolley is instantaneously at rest, and state when it reverses and in which direction.
Read the worked answer
Differentiate
v = ds/dt = 3t² − 18t + 24 = 3(t² − 6t + 8) = 3(t − 2)(t − 4)
a = dv/dt = 6t − 18
Rest
v = 0 at t = 2 s and t = 4 s (both inside 0 ≤ t ≤ 6).
Positions there: s(2) = 8 − 36 + 48 = 20 m; s(4) = 64 − 144 + 96 = 16 m.
Reversal
v > 0 for 0 ≤ t < 2 (moving away from O), so at t = 2 s the trolley reverses and travels back towards O.
v < 0 for 2 < t < 4, so at t = 4 s it reverses again and moves away from O.
After t = 4 s, v > 0 again. Two reversals in total.
2. Find the total distance travelled in 0 ≤ t ≤ 6, explain why it differs from the displacement, and decide whether the buffer at s = 40 m is reached.
Read the worked answer
Key positions
s(0) = 0, s(2) = 20, s(4) = 16, s(6) = 216 − 324 + 144 = 36 m.
Distance by interval (speed must be positive, so split at every reversal)
| Interval | Motion | Distance |
|---|---|---|
| 0 → 2 s | 0 m to 20 m | 20 m |
| 2 → 4 s | 20 m back to 16 m | 4 m |
| 4 → 6 s | 16 m to 36 m | 20 m |
Total distance = 20 + 4 + 20 = 44 m.
Displacement = s(6) − s(0) = 36 m, which is smaller because the 4 m of backward travel is cancelled in the displacement but still counts on the track.
Buffer check
s increases on [4, 6], so the greatest position reached during the test is s(6) = 36 m (the earlier peak was only 20 m). 36 m < 40 m, so the buffer is never contacted; the trolley finishes 4 m short of it.
3. Evaluate the trainee's and the foreman's statements, with numbers to support each judgement.
Read the worked answer
Trainee: 'it stopped at t = 3 s because a = 0'
At t = 3 s, a = 6(3) − 18 = 0, but v = 3(3 − 2)(3 − 4) = 3(1)(−1) = −3 m/s. The trolley is still moving at 3 m/s towards O. Zero acceleration means the velocity is momentarily not changing - here v has its minimum value of −3 m/s - not that the trolley is at rest. Rest requires v = 0, which happens only at t = 2 s and t = 4 s. The trainee has confused dv/dt = 0 with v = 0.
Foreman: 'distance = change in s = 36 m'
36 m is the displacement, not the distance. Because the trolley reverses at t = 2 s and t = 4 s, the motion must be split into three parts and the magnitudes added: 20 + 4 + 20 = 44 m. Using the net change understates the wear on the track and the trolley's travel by 8 m.
Both errors come from skipping the sign analysis of v; the correct method is to find v = 0 first, then treat each interval separately.