Identify a change of direction
A ball thrown up
A ball is thrown straight up, and up is the positive direction. The top graph shows its height h, in meters, after t seconds. The lower graph shows its velocity v, shaded green where v is positive and red where v is negative.
At t = 1 the tangent slopes up and the velocity is 10, so the ball moves up. The height curve is green where the height rises and red where it falls.
At t = 3 the tangent slopes down and the velocity is −10, so the ball moves down.
At t = 2 the tangent is flat and the velocity is 0. The velocity is positive just before t = 2 and negative just after it, so the ball turns around at the top.
A zero with a sign change
Drag the point past the peak and past the valley of the position graph.
At each of them the tangent is flat, so the velocity is 0. The velocity also changes sign there, so the particle changes direction.
A change of direction needs a change of sign. A zero of the velocity is the place to test, and the signs on both sides decide.
Notation
A particle changes direction at a time \(t_0\) when the velocity is 0 there and changes sign.
\[\begin{gathered} \cy{v}(t_0) = 0 \quad\text{and} \\ \text{the sign of}\ \cy{v}\ \text{changes at}\ t_0 \end{gathered}\]If v goes from positive to negative, the particle turns from the positive to the negative direction, and the position graph has a peak.
If v goes from negative to positive, the particle turns from the negative to the positive direction, and the position graph has a valley.
Find the zeros of v. Then compare the sign of v shortly before and shortly after each zero.
When it fails
At k = 1 the velocity crosses the axis twice, so the particle changes direction twice. As you lower k, the two zeros move together and join at t = 3.
At k = 0 the velocity is 0 at t = 3, but it is positive on both sides. The particle stops for an instant and moves on in the same direction.
Example 1: A velocity that crosses twice
A particle on a line has velocity \(v(t) = t^2 - 6t + 8\) meters per second, for \(t \ge 0\). Find when it changes direction.
Factor the velocity. It is 0 at \(t = 2\) and at \(t = 4\).
\[v(t) = (t - 2)(t - 4)\]The sign of v can change only at a zero, so test one time before, between and after the zeros.
\[\begin{gathered} v(1) = \cg{3}, \quad v(3) = \cr{-1} \\ v(5) = \cg{3} \end{gathered}\]At \(t = 2\) the sign goes from positive to negative, and at \(t = 4\) it goes from negative to positive. The particle changes direction at both times.
Example 2: A velocity that touches zero
A particle has position \(s(t) = t^4 - 8t^3 + 18t^2 - 16t\) meters after \(t\) seconds, for \(t \ge 0\). Find when it changes direction.
Differentiate and factor the velocity. It is 0 at \(t = 1\) and at \(t = 4\).
\[\begin{aligned} v(t) &= 4t^3 - 24t^2 + 36t - 16 \\ &= 4(t - 1)^2(t - 4) \end{aligned}\]A tempting wrong move: The velocity is 0 at \(t = 1\) and at \(t = 4\), so the particle changes direction at both times.
The graph of v touches the axis at t = 1 and returns below it. The position graph stays red through t = 1, so the particle does not turn.
Test one time before, between and after the zeros.
\[\begin{gathered} v(0.5) = \cr{-3.5}, \quad v(2) = \cr{-8} \\ v(5) = \cg{64} \end{gathered}\]At \(t = 1\) the velocity is negative on both sides, so the particle stops for an instant and continues in the same direction. At \(t = 4\) the sign goes from negative to positive, so the particle changes direction only at \(t = 4\).
Practice
A cart has velocity \(v(t) = (t - 2)(t - 5)^2\) meters per second, for \(t \ge 0\). Find when it changes direction.
Drag the point across each place where the graph of v meets the axis. Compare the color of the shaded region on the two sides.
Solution
The zeros of \(v\) are \(t = 2\) and \(t = 5\). Test one time in each interval.
\[\begin{gathered} v(1) = \cr{-16}, \quad v(3) = \cg{4} \\ v(6) = \cg{4} \end{gathered}\]At \(t = 2\) the sign goes from negative to positive, so the cart changes direction. At \(t = 5\) the velocity is positive on both sides, so it does not.
Questions
These questions use a particle with velocity \(v(t) = \tfrac{1}{8}(t - 1)(t - 3)^2(t - 5)\) meters per second. The top graph shows its position \(s\), and the lower graph shows \(v\).
At \(t = 3\) the graph of \(v\) touches the axis from below. What does the particle do?
- It stops for an instant and continues in the negative direction.
- It changes direction, because the velocity is 0.
- It stays at rest from then on.
- It continues in the positive direction.
Answer
It stops for an instant and continues in the negative direction.
At \(t = 5\) the graph of \(v\) crosses the axis. What does the particle do?
- It changes from the negative direction to the positive direction.
- It changes from the positive direction to the negative direction.
- It continues in the negative direction.
Answer
It changes from the negative direction to the positive direction.
A particle has position \(s(t) = t^3 - 3t^2 + 3t\) meters. Does it ever change direction?
- No. The velocity is 0 at \(t = 1\) but is positive on both sides.
- Yes, at \(t = 1\), because \(v(1) = 0\).
- No, because the velocity is never 0.
- Yes, at \(t = 0\) and \(t = 1\).
Answer
No. The velocity is 0 at \(t = 1\) but is positive on both sides.
Summary
You find the zeros of the velocity, then compare the sign of the velocity shortly before and shortly after each zero.
A particle turns only where the velocity changes sign, so a zero where the graph touches the axis is a stop, not a turn.
You use this on a position function or a velocity graph to find every time the particle reverses.
This is one of five skills in topic 4.2 of Which & Why: AP Calculus.