Find rest times and direction of motion
A particle on a line
A particle moves on a line. The graph shows its velocity v, in meters per second, after t seconds. The line below the graph shows where the particle is, and the arrows above the line trace its path.
Drag the orange edge to the right. Where v is positive, the graph is shaded green and the particle moves to the right, as at t = 1, where v = 12.
Where v is negative, the graph is shaded red and the particle moves to the left, as at t = 5, where v = −4.
The graph crosses the axis at t = 3 and at t = 7. Press Mark the times where v = 0 to show them with dashed lines. At these times the velocity is 0, so the particle is at rest for an instant.
Zeros split time into intervals
Without numbers, the times where the graph crosses the axis still divide the time axis into intervals. The labels name the direction on each one.
Drag the orange edge across one interval. The color of the graph stays the same inside it and changes only at a crossing.
Because v has one sign on each interval, its sign at one chosen time, a test time, gives the direction for the whole interval.
Notation
The sign of the velocity gives the direction. If \(v > 0\), the particle moves in the positive direction. If \(v < 0\), it moves in the negative direction.
\[v > 0\ \text{(right)}, \qquad v < 0\ \text{(left)}\]The particle is at rest when its velocity is 0, so the rest times are the solutions of this equation.
\[\cy{v(t)} = 0\]The rest times split the time axis into intervals. A continuous velocity changes sign only where it is 0, so it keeps one sign on each interval.
Choose one test time in each interval and find the sign of \(v\) there.
Example 1: A skater
A skater has position \(s(t) = t^2 - 6t + 5\) meters after \(t\) seconds, for \(t \ge 0\). Find when the skater is at rest and the direction of motion.
The velocity is the derivative of the position.
\[v(t) = 2t - 6\]Set the velocity equal to 0. The skater is at rest at \(t = 3\).
\[2t - 6 = 0 \ \Rightarrow\ t = 3\]Test one time before 3. The velocity is negative, so the skater moves in the negative direction for \(0 \le t < 3\).
\[v(1) = \cr{-4}\]Test one time after 3. The velocity is positive, so the skater moves in the positive direction for \(t > 3\).
\[v(5) = \cg{4}\]
Example 2: A bead on a wire
A bead has position \(s(t) = 2t^3 - 21t^2 + 60t\) centimeters after \(t\) seconds, for \(t \ge 0\). Find when the bead is at rest and the direction of motion.
Differentiate and factor the velocity. The rest times are \(t = 2\) and \(t = 5\).
\[\begin{aligned} v(t) &= 6t^2 - 42t + 60 \\ &= 6(t - 2)(t - 5) \end{aligned}\]A tempting wrong move: At \(t = 0\) the velocity is 60, which is positive, so the bead moves in the positive direction for all \(t\).
\[v(0) = 60 > 0\]The graph is below the axis between t = 2 and t = 5, so the bead moves left there. The red dot shows where the bead would be if it never turned, and the orange bead is to its left. One test time describes only its own interval.
Test one time in each of the three intervals.
\[\begin{gathered} v(1) = \cg{24}, \quad v(3) = \cr{-12} \\ v(6) = \cg{24} \end{gathered}\]The bead moves in the positive direction for \(0 \le t < 2\) and for \(t > 5\). It moves in the negative direction for \(2 < t < 5\).
Practice
A particle has position \(s(t) = t^3 - 9t^2 + 15t\) meters after \(t\) seconds, for \(t \ge 0\). Find when it is at rest and when it moves in the negative direction.
Factor the velocity to find its zeros. Then drag the edge and see where the graph is red.
Solution
Differentiate and factor. The particle is at rest at \(t = 1\) and \(t = 5\).
\[\begin{aligned} v(t) &= 3t^2 - 18t + 15 \\ &= 3(t - 1)(t - 5) \end{aligned}\]Test one time in each interval.
\[\begin{gathered} v(0) = \cg{15}, \quad v(2) = \cr{-9} \\ v(6) = \cg{15} \end{gathered}\]The velocity is negative only on the middle interval, so the particle moves in the negative direction for \(1 < t < 5\).
Questions
These questions use a particle with velocity \(v(t) = \tfrac{1}{4}(t - 1)(t - 4)(t - 6)\) meters per second. The graph shows \(v\), and the line below it shows the particle's position.
At \(t = 3\) the graph is above the axis and slopes down. Which statement is correct?
- The particle moves in the positive direction.
- The particle moves in the negative direction, because the graph slopes down.
- The particle is at rest.
Answer
The particle moves in the positive direction.
At \(t = 5\) the graph is below the axis, between the rest times \(t = 4\) and \(t = 6\). Which statement is correct?
- The particle moves in the negative direction.
- The particle moves in the positive direction, because it has passed a rest time.
- The particle is at rest for the whole interval.
Answer
The particle moves in the negative direction.
A particle has position \(s(t) = t^3 - 3t^2 - 45t\) centimeters, for \(t \ge 0\). Which statement is correct?
- It is at rest at \(t = 5\). It moves in the negative direction for \(0 \le t < 5\) and in the positive direction for \(t > 5\).
- It is at rest at \(t = -3\) and \(t = 5\).
- It is at rest at \(t = 5\). It moves in the positive direction for \(0 \le t < 5\) and in the negative direction for \(t > 5\).
- It is at rest at \(t = 5\) and moves in the positive direction at every other time.
Answer
It is at rest at \(t = 5\). It moves in the negative direction for \(0 \le t < 5\) and in the positive direction for \(t > 5\).
Summary
You solve v(t) = 0 for the rest times, then test the sign of v in each interval between them.
This works because the velocity can change sign only at a zero, so one test time gives the direction on the whole interval.
You use it on a position function or a velocity graph to say when a particle rests and which way it moves.
This is one of five skills in topic 4.2 of Which & Why: AP Calculus.