Position, velocity and acceleration

When you reach for it

A paddleboard on a straight course, a bead on a wire, a piston in a cylinder: each moves along a single line, and its whole motion is recorded by one function, its position s(t) at time t. Questions about the motion (“is it moving?”, “which way?”, “when does it turn back?”, “is it gaining speed?”, “how far did it go?”) all become questions about the signs of s' and s''.

The derivative as a rate of change, with its units and sign, is The derivative as a rate of change; the second derivative as acceleration is in Second and higher derivatives, where Matus’s quadcopter already computes height, velocity and acceleration. This lesson reads those derivatives as motion, and the tool it leans on is the sign chart of Increasing and decreasing from the derivative, now drawn for velocity.

The procedure

Day 1: rest, direction and turns

Position, velocity, acceleration. Put a number line along the path, choose an origin and a positive direction (call it “right”, or east). The position s(t) is the signed number on that line where the object sits at time t. Its velocity is the instantaneous rate of change of position, and its acceleration is the rate of change of velocity:

Units follow the difference quotients. If s is in meters and t in minutes, then \dfrac{s(t+h) - s(t)}{h} is in meters per minute, so v is in meters per minute; and \dfrac{v(t+h) - v(t)}{h} is in meters per minute per minute, written \text{m/min}^2.

Which way it moves. On an interval where v(t) > 0, the test in Increasing and decreasing from the derivative says s is increasing, so the object moves right. Where v(t) < 0, s is decreasing and it moves left. At an instant c with v(c) = 0 the object is at rest for that instant.

A turn is a sign change. The object changes direction at c when v changes sign at c: positive just before and negative just after, or the reverse. For a continuous velocity this forces v(c) = 0, by the intermediate value theorem. The converse fails: v(c) = 0 alone does not make a turn, because v can touch 0 and keep its sign (Example 2). So a turn is found exactly as a change of direction of any function is found, with a sign chart of v: split the time line at the zeros of v and at any time where v does not exist, test one value in each piece, and look for neighboring pieces with opposite signs.

Where the turns put it. Between two consecutive turns, s is increasing or decreasing throughout, so the object’s extreme positions on a time interval are at its turns or at the two ends. The positions at the turns are the values of s worth computing.

Day 2: speed, speeding up and distance

Speed. Velocity carries a direction in its sign; speed is how fast, with no direction: speed = \lvert v(t) \rvert. A velocity of -12 meters per minute is a speed of 12 meters per minute, moving left.

Speeding up and slowing down. The object is speeding up on an interval if its speed is increasing there, and slowing down if its speed is decreasing.

> On an interval where v(t)\,a(t) > 0 (velocity and acceleration of the same sign) the object is speeding up. Where v(t)\,a(t) < 0 (opposite signs) it is slowing down.

The reason is one line of the chain rule (The chain rule):

If va > 0 on an interval, then v^2 is increasing there. Since speed is \displaystyle \lvert v \rvert = \sqrt{v^2} and the square root is increasing on [0, \infty), the speed is increasing too. If va < 0, the same argument gives v^2 decreasing, and so the speed decreasing. Where v \ne 0 this also gives the rate of change of speed directly: \dfrac{d}{dt}\lvert v \rvert = \dfrac{v\,a}{\lvert v \rvert}, which has the sign of va.

So the sign of a alone does not say whether the object speeds up. A negative acceleration makes the velocity decrease; if the velocity is already negative, it becomes more negative, and the object moving left goes faster.

Displacement and distance. Over a time interval [t_0, t_1], the displacement is the net change in position, s(t_1) - s(t_0), which can be negative. The total distance traveled counts every meter covered, whichever way. Cut [t_0, t_1] at the turns. On each piece the object moves one way only, so the distance covered on a piece [t_i, t_{i+1}] is \lvert s(t_{i+1}) - s(t_i) \rvert, and

Displacement and distance agree exactly when there is no turn inside the interval. When only the velocity is known and not the position, both come from integrals of the velocity, which is a later lesson: Position, displacement and distance from velocity.

Reading a graph of velocity. On a graph of v against t: the height gives the velocity (above the axis, moving right; below, moving left); the distance from the axis is the speed; the slope is the acceleration. A crossing of the axis is a turn, a touch without crossing is not. The board speeds up where the graph moves away from the axis and slows down where it moves toward it, which is the rule va > 0 read off a picture.

Worked examples

Boundaries

• The rules need a differentiable position. Velocity is s'; where s has a corner, v does not exist and that time goes on the sign chart as a split point. Where v has a corner, as in Figure 3, a does not exist there.
• v(c) = 0 is necessary for a turn, not sufficient. A continuous velocity can only change sign through 0, but a zero with no sign change (an even power of t - c) is a pause, not a turn.
• A rest at an end of the time interval is not a turn. Nothing is known about the motion outside the interval.
• Speeding up is a statement about speed, not velocity. It needs the signs of both v and a; the rule says nothing at an instant where v = 0, where the speed reaches its lowest value.
• The distance formula needs every turn. Skip a sign change of v and the sum undercounts: the piece that contains the missed turn reports only its net change.

Where this goes next

When only the velocity is known, displacement and distance come from integrating v and \lvert v \rvert: Position, displacement and distance from velocity. In the plane, velocity becomes a vector and speed its length, in Vector valued functions and motion in the plane and Arc length and speed on parametric and polar curves. The same pairing of first and second derivatives that decides speeding up also decides the bend of a graph, in Concavity and inflection points.

Exercises

Day 1: rest, direction and turns

For Exercises 1 to 6, s(t) is the position of a particle on a line. Find v(t) and a(t), the times when the particle is at rest, the intervals on which it moves right and left, and every time it changes direction.

1. s(t) = t^3 - 9t^2 + 15t, t \ge 0
2. s(t) = t^4 - 8t^2, t \ge 0
3. s(t) = t^4 - 4t^3, -1 \le t \le 4
4. s(t) = t - 2\sin t, 0 \le t \le 2\pi
5. s(t) = \dfrac{t}{t^2 + 9}, t \ge 0
6. s(t) = (t^2 - 4)^3, -3 \le t \le 3
7. Keola’s position is s(t) = 3t^2 - t^3 meters east of a buoy, t minutes after he starts, for 0 \le t \le 4. Find v(3) and a(3) with units, and say in words what each tells you.

Day 1: reasoning

8. Iolana says a particle with velocity v(t) = (t - 3)^2 turns around at t = 3 because v(3) = 0. Explain what is wrong.
9. A particle’s velocity is continuous on [0, 5] with v(0) = 2 and v(5) = -1. Explain why it is at rest at least once, and why it must reverse direction somewhere in (0, 5).
10. Give a position function and an instant at which the particle is at rest but its acceleration is not 0, and one whose acceleration is 0 at every instant though it is never at rest.
11. Prove: if v is differentiable, v(c) = 0 and a(c) \ne 0, then the particle changes direction at c.

Day 2: speeding up and slowing down

For Exercises 12 to 15, find the intervals on which the particle is speeding up and on which it is slowing down.

12. s(t) = t^3 - 6t^2 + 9t, t \ge 0
13. s(t) = 2t^3 - 3t^2 - 12t, t \ge 0
14. \displaystyle s(t) = 5\sin\left(\frac{\pi t}{2}\right), 0 \le t \le 4. Also find the greatest speed and when it occurs.
15. s(t) = t^2 e^{-t}, t \ge 0

Day 2: distance and displacement

16. For the particle of Exercise 1, find the displacement and the total distance traveled on [0, 6].
17. For the particle of Exercise 2, find the displacement and the total distance traveled on [0, 3].
18. For the particle of Exercise 4, find the displacement and the total distance traveled on [0, 2\pi].
19. Iolana’s position is s(t) = t^3 - 15t^2 + 63t meters from the dock, t minutes after she starts, for 0 \le t \le 9. Find her displacement, the total distance she paddles, and her greatest speed.
20. A particle has position s(t) = \ln(t^2 + 1) - t for 0 \le t \le 3. Show that it never changes direction, and find the total distance it travels.

Day 2: reading a velocity graph

21. Using Figure 3, give Keola’s acceleration on each straight piece, and the times at which his acceleration does not exist.
22. Using Figure 3, find every time at which Keola’s speed is 1 meter per second.

Day 2: reasoning

23. A particle has position s(t) = (t - 2)^3 on [0, 4]. Akoni finds the distance by splitting at t = 2, where v = 0, and adding \lvert s(2) - s(0) \rvert + \lvert s(4) - s(2) \rvert. Iolana computes \lvert s(4) - s(0) \rvert. Who is right?
24. Prove: on an interval where v(t)\,a(t) < 0, the particle is slowing down.
25. Keola says, “the acceleration is positive, so the particle is speeding up.” Give a position function that shows he is wrong, and state what is true.