Wemazi

Topic 4.2

Decide when speed increases or decreases

Builds on identifying a change of direction. 6 minutes.

A skateboard on a path

A skateboard rolls on a straight path. The top graph shows its velocity v, in meters per second, and the lower graph shows its acceleration a, which is the slope of v. Move the slider. The two orange bars show v and a at that time.

The faint dashed curve shows the speed, which is how fast the board moves in either direction. It is green where the speed rises and red where it falls.

At t = 1 the velocity is 8, so the board moves forward. The acceleration is −4, so the velocity falls and the board slows down.

At t = 5 the velocity is −8, so the board moves backward. The acceleration is still −4, so the velocity becomes more negative and the board moves backward faster.

The speed falls to 0 at t = 3, where the board stops and turns around, and it rises after that.

Compare the two signs

Drag the point. The acceleration is negative at every time. The speed curve is red where v is positive, so the speed falls there, and green where v is negative, so the speed rises there.

The speed rises when v and a have the same sign and falls when their signs are opposite. The acceleration pushes either with the motion or against it.

Notation

  1. The speed is the size of the velocity, so it is never negative.

    \[\text{speed} = \cy{|v(t)|}\]

  2. If v and a have the same sign, the speed increases, because the acceleration pushes the particle in the direction it already moves.

    \[v\,a > 0\]

  3. If v and a have opposite signs, the speed decreases, because the acceleration pushes against the motion.

    \[v\,a < 0\]

  4. The sign of a alone does not decide. A negative a slows a particle that moves to the right, and it speeds up a particle that moves to the left.

  5. At a time when v = 0 or a = 0, neither rule applies. Find the zeros of both and test one time in each interval between them.

Example 1: A glider

A glider has velocity \(v(t) = t^2 - 9\) meters per second, for \(t \ge 0\). Decide when its speed increases and when it decreases.

  1. The acceleration is the derivative of the velocity, so it is positive for all t > 0.

    \[a(t) = 2t\]

  2. The velocity is 0 when t² = 9, which is at t = 3 because t ≥ 0. Test one time on each side.

    \[v(1) = \cr{-8}, \quad v(4) = \cg{7}\]

  3. For \(0 < t < 3\), \(v\) is negative and \(a\) is positive, so the speed decreases.

  4. For \(t > 3\), \(v\) and \(a\) are both positive, so the speed increases.

Example 2: A particle with four intervals

A particle has position \(s(t) = t^3 - 12t^2 + 36t\) meters after \(t\) seconds, for \(t \ge 0\). Decide when its speed increases and when it decreases.

  1. Differentiate twice and factor. The velocity is 0 at t = 2 and t = 6, and the acceleration is 0 at t = 4.

    \[\begin{gathered} v(t) = 3(t - 2)(t - 6) \\ a(t) = 6(t - 4) \end{gathered}\]

  2. A tempting wrong move: The acceleration is negative for \(0 < t < 4\), so the particle slows down on the whole interval.

    The speed curve is green between t = 2 and t = 4, so the speed rises there. The velocity is also negative there, so the negative acceleration pushes with the motion.

  3. The three zeros split the time axis into four intervals. Test one time in each.

    \[\begin{array}{c|cccc} t & 1 & 3 & 5 & 7 \\ \hline v & 15 & -9 & -9 & 15 \\ a & -18 & -6 & 6 & 18 \end{array}\]

  4. The signs are opposite for \(0 < t < 2\) and for \(4 < t < 6\), so the speed decreases. They are the same for \(2 < t < 4\) and for \(t > 6\), so the speed increases.

Practice

A particle has position \(s(t) = 2t^3 - 15t^2 + 36t\) centimeters after \(t\) seconds. Decide whether its speed increases or decreases at \(t = 1\) and at \(t = 2.2\).

Find v and a at each time. Then compare their signs, which the two orange bars show.

Solution

  1. Differentiate twice.

    \[\begin{aligned} v(t) &= 6t^2 - 30t + 36 \\ &= 6(t - 2)(t - 3) \\ a(t) &= 12t - 30 \end{aligned}\]

  2. At \(t = 1\), \(v\) is positive and \(a\) is negative, so the speed decreases.

    \[v(1) = \cg{12}, \qquad a(1) = \cr{-18}\]

  3. At \(t = 2.2\), \(v\) and \(a\) are both negative, so the speed increases.

    \[\begin{gathered} v(2.2) = \cr{-0.96} \\ a(2.2) = \cr{-3.6} \end{gathered}\]

Questions

These questions use a particle with velocity \(v(t) = -t^2 + 8t - 12\) meters per second. The graphs show \(v\) and its acceleration \(a = -2t + 8\).

  1. At \(t = 3\) the velocity is 3 and the acceleration is 2. What happens to the speed?

    • It increases, because \(v\) and \(a\) are both positive.
    • It decreases, because the graph of v bends downward.
    • It stays constant, because the velocity is small.
    Answer

    It increases, because \(v\) and \(a\) are both positive.

  2. At \(t = 7\) the velocity is \(-5\) and the acceleration is \(-6\). What happens to the speed?

    • It increases, because \(v\) and \(a\) are both negative.
    • It decreases, because the acceleration is negative.
    • It decreases, because the velocity is negative.
    • It is 0, because the particle moves backward.
    Answer

    It increases, because \(v\) and \(a\) are both negative.

  3. A particle has position \(s(t) = t^3 - 6t^2\) meters. What happens to its speed at \(t = 1\)?

    • It increases, because \(v(1) = -9\) and \(a(1) = -6\) have the same sign.
    • It decreases, because \(a(1)\) is negative.
    • It decreases, because \(v(1)\) is negative.
    • It increases, because \(s(1) = -5\) is negative.
    Answer

    It increases, because \(v(1) = -9\) and \(a(1) = -6\) have the same sign.

Summary

The speed of a particle increases when its velocity and acceleration have the same sign, because the acceleration pushes with the motion.

The speed decreases when the signs are opposite, because the acceleration pushes against the motion.

To decide when a particle speeds up or slows down, compare the signs of v and a on each interval between their zeros.

This is one of five skills in topic 4.2 of Which & Why: AP Calculus.