Velocity Sign Changes: Displacement Versus Distance
A particle moves along a straight line. Its velocity, in metres per second, is modelled by v(t)=3t²−12t+9 for 0≤t≤4, where t is in seconds. At t=0 the particle is at the origin, and displacement is measured positive to the right. A student integrates v(t) from 0 to 4 and obtains 4 m, then concludes that the particle has travelled 4 m in total. A second student integrates the speed instead and gets a different value, but neither student has drawn a velocity–time sketch or checked where v changes sign. The examiner asks for the displacement at t=4, the total distance travelled, and the time at which the particle returns to its starting point. The student must also explain the difference between displacement and distance in this context, using the sign of v. Answers for distance and displacement should be exact, and the reasoning must refer to the intervals between the times when v=0. The use of a calculator is allowed for checking, but the integrals must be evaluated exactly.
1. Find the times in 0≤t≤4 when v(t)=0 and determine the sign of v on each subinterval.
Read the worked answer
Factorise the velocity:
v(t)=3t²−12t+9=3(t²−4t+3)=3(t−1)(t−3).
So v=0 at t=1 and t=3. On the intervals between these times:
- 0≤t<1: choose t=0, v=9>0, so v is positive.
- 1<t<3: choose t=2, v=12−24+9=−3<0, so v is negative.
- 3<t≤4: choose t=4, v=48−48+9=9>0, so v is positive.
Thus the particle moves to the right on [0,1), to the left on (1,3), and to the right again on (3,4].
2. Find the displacement at t=4 and the total distance travelled over 0≤t≤4. Show the definite integrals.
Read the worked answer
Integrate velocity to get displacement, using s(0)=0:
s(t)=∫(3t²−12t+9)dt=t³−6t²+9t+C, and C=0.
So the displacement at t=4 is s(4)=64−96+36=4 m. The positive sign means the particle is 4 m to the right of the origin.
Distance is the integral of speed, so split at the sign changes:
Distance=∫₀¹ v dt − ∫₁³ v dt + ∫₃⁴ v dt.
Let F(t)=t³−6t²+9t. Then F(0)=0, F(1)=1−6+9=4, F(3)=27−54+27=0, F(4)=4.
Distance=|4−0|+|0−4|+|4−0|=4+4+4=12 m.
The displacement is only 4 m because the leftward motion cancels part of the rightward motion; the distance is larger because it accumulates every metre travelled regardless of direction.
3. Find the time at which the particle returns to its starting point, and explain why the displacement is zero there while the distance travelled is not.
Read the worked answer
The displacement function is
s(t)=t³−6t²+9t=t(t²−6t+9)=t(t−3)².
So s(t)=0 at t=0 and t=3. The particle returns to its starting point at t=3 s.
At t=3 the net displacement from the origin is zero, but the particle has not stayed still. It moved right by ∫₀¹ v dt=4 m, then moved left by ∫₁³ v dt=−4 m. The distance travelled by t=3 is therefore 4+4=8 m, even though the displacement is 0. Displacement is the net change in position and can cancel; distance is the total of the magnitudes of the movements and cannot be negative.