Speed, Velocity, and the Particle That Turns Around
A particle moves along a straight line with velocity v(t)=t^2−6t+8 metres per second for 0≤t≤5, where t is in seconds. Its initial displacement is s(0)=0. A student computes the acceleration a(t)=2t−6 and observes that a(t)>0 for t>3, so concludes that the particle's speed is increasing for all t>3. The student also calculates the displacement over the full five seconds as 6.67 m and claims the particle has travelled 6.67 m in total. The teacher points out that speed and velocity are not the same, and that displacement and total distance are not the same. The particle changes direction when v(t)=0. The class is asked to describe the motion correctly, including when the particle is momentarily at rest, when its speed is increasing, and the total distance travelled over 0≤t≤5.
1. Find all times in 0≤t≤5 when the particle is at rest, and make a sign chart for v(t). On which intervals is the particle moving in the positive direction?
Read the worked answer
v(t)=t^2−6t+8=(t−2)(t−4). Thus v(t)=0 at t=2 and t=4.
Sign chart:
- 0≤t<2: choose t=0, v=8>0 (positive direction).
- 2<t<4: choose t=3, v=9−18+8=−1<0 (negative direction).
- 4<t≤5: choose t=5, v=25−30+8=3>0 (positive direction).
The particle is at rest instantaneously at t=2 and t=4. It moves in the positive direction on [0,2) and (4,5].
2. The student says speed is increasing for all t>3 because a(t)>0. Determine the correct intervals where speed is increasing. Explain the flaw in the student's reasoning.
Read the worked answer
Speed is |v(t)|. When v ≠ 0, the derivative of speed has the sign of v(t)a(t), so speed increases when velocity and acceleration have the same sign.
a(t)=2t−6=2(t−3). Thus a<0 for t<3 and a>0 for t>3.
Check intervals:
- 0<t<2: v>0, a<0 → speed decreasing.
- 2<t<3: v<0, a<0 → speed increasing.
- 3<t<4: v<0, a>0 → speed decreasing.
- 4<t<5: v>0, a>0 → speed increasing.
So speed is increasing on (2,3) and (4,5), not for all t>3. The student's error is assuming positive acceleration always means increasing speed. For 3<t<4 the particle is still moving in the negative direction while acceleration is positive, so the acceleration is slowing it down (speed decreasing).
3. Calculate the total distance travelled over 0≤t≤5 and compare it with the displacement. Show the split at the turning points and explain why the two quantities differ.
Read the worked answer
Displacement is
∫0^5 v(t)dt = [t^3/3 − 3t^2 + 8t]_0^5 = 125/3 − 75 + 40 = 20/3 ≈ 6.67 m.
For total distance, split at the turning points t=2 and t=4. Position values: s(0)=0, s(2)=8/3−12+16=20/3, s(4)=64/3−48+32=16/3, s(5)=125/3−75+40=20/3.
Distance = |s(2)−s(0)| + |s(4)−s(2)| + |s(5)−s(4)| = 20/3 + |16/3−20/3| + |20/3−16/3| = 20/3 + 4/3 + 4/3 = 28/3 ≈ 9.33 m.
The displacement is about 6.67 m, while the total distance is about 9.33 m. They differ because the particle turns around at t=2 and t=4; the negative and positive displacements partly cancel in the displacement calculation, but distance adds the magnitudes of each segment.