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Calculus IChapter 3: DerivativesSection 3-2

Interpretation of the Derivative — Question 7

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Question 7

The position of a particle moving along a straight line is given by the function s(t)=t3−6t2+9ts(t) = t^3 - 6t^2 + 9t where s(t)s(t) is in meters and t≥0t\ge0 is in seconds.

  • (a) Find the velocity function v(t)v(t).

  • (b) At what time(s) is the particle at rest?

  • (c) Determine the direction of motion (moving forward or backward) on the intervals determined by part (b).

  • (d) Is the particle speeding up or slowing down at t=1t = 1? Justify using acceleration.

Original worksheet page 1: question and worked solution for 3-2-007
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Question 7 - Solution

Velocity and rest times.

v(t)=3t2−12t+9=3(t−1)(t−3),a(t)=6t−12.v(t)=3t^2-12t+9=3(t-1)(t-3),\qquad a(t)=6t-12.

Hence the particle is at rest at t=1,3\boxed{t=1,3}.

For t≥0t\ge0, it moves forward on [0,1)∪(3,∞)[0,1)\cup(3,\infty) and backward on (1,3)(1,3).

Behavior at t=1t=1. Here v(1)=0v(1)=0 and a(1)=−6a(1)=-6. Just before 11, velocity and acceleration have opposite signs, so speed decreases. Just after 11, both are negative, so speed increases.

The speed |v(t)||v(t)| has left derivative −6-6 and right derivative 66 at 11; it is not differentiable there.

At t=1 it is at rest and reverses direction.\boxed{\text{At }t=1\text{ it is at rest and reverses direction.}}

It slows down immediately before that instant and speeds up immediately afterward.

Original worksheet page 2: question and worked solution for 3-2-007

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