Grade by Grade
B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Speeding Up, Slowing Down, and Total Distance

B2.1About 65 minutes
Learning goals
  • I can decide when a moving object is speeding up or slowing down from the signs of v and a.
  • I can make a sign chart for velocity and acceleration.
  • I can find the total distance travelled and tell it apart from displacement.
  • I can calculate average velocity and average speed.

Words to know

speeding up
The speed |v| is increasing. This happens when v and a have the same sign.
slowing down
The speed |v| is decreasing. This happens when v and a have opposite signs.
turning point (of motion)
A time when the velocity changes sign, so the object reverses direction.
total distance travelled
The sum of the sizes of all the displacements between turning points.
average velocity
Displacement ÷ elapsed time.
average speed
Total distance ÷ elapsed time.

Is negative acceleration always braking?

Picture a car that is reversing out of a driveway faster and faster. Call forward positive. The car's velocity is negative (it moves backward) and its velocity is getting more negative, so its acceleration is negative too. Yet the car is speeding up. So "negative acceleration" and "slowing down" are not the same thing. This lesson gives you a reliable test, and shows the difference between displacement and total distance travelled.

Speeding up or slowing down?

Speed is |v|. The speed increases when the velocity moves away from 0, and decreases when it moves toward 0.

  • If v > 0 and a > 0, the velocity is positive and growing: speeding up.
  • If v < 0 and a < 0, the velocity is negative and becoming more negative: speeding up.
  • If v and a have opposite signs, the velocity is moving toward 0: slowing down.

Rule: an object is speeding up when v and a have the same sign and slowing down when they have opposite signs.

To use the rule, find where v = 0 and where a = 0, mark these times on a time line, and test the sign of each in every interval (a sign chart, like the ones you used in Curve Sketching: Connecting f, f′ and f″).

Worked example

Example 1: Speeding up and slowing down

For the particle from the last lesson, s(t) = t³ − 6t² + 9t (m, s), find the intervals on 0 ≤ t ≤ 4 when it is speeding up and when it is slowing down.

  1. v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3), zero at t = 1 and t = 3.
  2. a(t) = 6t − 12 = 6(t − 2), zero at t = 2.
  3. These split [0, 4] into (0, 1), (1, 2), (2, 3) and (3, 4). Test one time in each: at t = 0.5, v = 3.75 > 0 and a = −9 < 0. At t = 1.5, v = −2.25 < 0 and a = −3 < 0. At t = 2.5, v = −2.25 < 0 and a = 3 > 0. At t = 3.5, v = 3.75 > 0 and a = 9 > 0.
  4. Same signs mean speeding up; opposite signs mean slowing down.

Slowing down on (0, 1) and (2, 3). Speeding up on (1, 2) and (3, 4).

t01234v = 3(t − 1)(t − 3)+−−+a = 6(t − 2)−−++speed is…slowing downspeeding upslowing downspeeding up
The sign chart for Example 1. Each column is a time interval. When the signs of v and a match, the particle speeds up; when they differ, it slows down.
1234−30369t (s)m/sslowingspeeding upslowingspeeding upspeed |v|velocity v
The speed |v(t)| (solid) for Example 1. Where v (dashed) is negative, the speed graph is its reflection above the axis. Speed falls to 0 at t = 1 and t = 3 (the turning points) and has a local high at t = 2, where the acceleration is 0.

Displacement versus total distance

Displacement only compares where the object ends with where it starts. If it goes out and back, the displacement can be small even though the object travelled a long way.

To find the total distance travelled from t = a to t = b:

  1. Find every time in (a, b) when the object turns around (v = 0 and v changes sign).
  2. Find the position at a, at each turning time, and at b.
  3. Add up the sizes of the changes in position between consecutive times.

Two useful averages:

  • average velocity = displacement ÷ elapsed time (can be negative),
  • average speed = total distance ÷ elapsed time (never negative).
Worked example

Example 2: Distance and displacement

s(t) = t³ − 9t² + 15t metres, 0 ≤ t ≤ 6 s. Find the displacement, the total distance travelled, the average velocity and the average speed.

  1. v(t) = 3t² − 18t + 15 = 3(t − 1)(t − 5). It changes sign at t = 1 and t = 5, so the particle turns around at those times.
  2. Positions: s(0) = 0; s(1) = 1 − 9 + 15 = 7; s(5) = 125 − 225 + 75 = −25; s(6) = 216 − 324 + 90 = −18.
  3. Displacement = s(6) − s(0) = −18 m.
  4. Distance = |7 − 0| + |−25 − 7| + |−18 − (−25)| = 7 + 32 + 7 = 46 m.
  5. Average velocity = −18 ÷ 6 = −3 m/s. Average speed = 46 ÷ 6 ≈ 7.67 m/s.

Displacement −18 m, total distance 46 m, average velocity −3 m/s, average speed about 7.67 m/s.

Worked example

Example 3: The thrown ball, again

For the ball with h(t) = −4.9t² + 14.7t + 1.5 (up is positive), when is it speeding up and when slowing down? What total distance does it travel before landing at t ≈ 3.099 s?

  1. v(t) = −9.8t + 14.7 and a(t) = −9.8 (always negative).
  2. For 0 ≤ t < 1.5, v > 0 and a < 0: opposite signs, so the ball slows down on the way up.
  3. For 1.5 < t ≤ 3.099, v < 0 and a < 0: same signs, so it speeds up on the way down, even though a is negative the whole time.
  4. Distance: up from 1.5 m to 12.525 m (11.025 m), then down from 12.525 m to 0 (12.525 m). Total = 23.55 m.

Slowing down for 0 ≤ t < 1.5 s, speeding up after. Total distance about 23.6 m.

Worked example

Example 4: Checking every interval

s(t) = 2t³ − 15t² + 36t (m, s), t ≥ 0. When is the object speeding up?

  1. v(t) = 6t² − 30t + 36 = 6(t − 2)(t − 3). a(t) = 12t − 30 = 6(2t − 5), zero at t = 2.5.
  2. Interval (0, 2): v > 0, a < 0, slowing. Interval (2, 2.5): v < 0, a < 0, speeding up.
  3. Interval (2.5, 3): v < 0, a > 0, slowing. Interval (3, ∞): v > 0, a > 0, speeding up.

Speeding up for 2 < t < 2.5 and t > 3.

Just for fun

A sign chart is the only chart where "plus times minus" means hit the brakes, and "minus times minus" means floor it.

Hands-on activity

Up the ramp and back

You need: a long board or a stiff piece of cardboard propped on a few books to make a gentle ramp, a toy car or a ball, and a ruler.

  1. Call up the ramp the positive direction.
  2. Give the car a gentle push up the ramp from the bottom (not hard enough to fly off the top). Watch it slow down, stop, and roll back.
  3. On paper, draw a time line with three stages: going up, the top, coming down. For each stage write the sign of v, the sign of a, and whether the car is speeding up or slowing down.
  4. Now make down the ramp the positive direction and redo the signs.

What to look for: with up positive, a is negative the whole time (gravity pulls the car back down the slope). Going up: v +, a −, slowing. At the top: v = 0. Coming down: v −, a −, speeding up. With down positive, every sign flips, but the speeding-up and slowing-down verdicts stay the same, because they depend only on whether the signs match.

Your turn

1

v(2) = 5 m/s and a(2) = −3 m/s². Is the object speeding up or slowing down at t = 2?

2

v(4) = −6 m/s and a(4) = −2 m/s². Is the object speeding up or slowing down at t = 4?

3

s(t) = t² − 8t + 7 (m, s). When is the object speeding up?

4

For s(t) = t² − 8t + 7 on 0 ≤ t ≤ 6, find the displacement and the total distance travelled.

5

s(t) = t³ − 12t (m, s), t ≥ 0. Find the intervals when it is speeding up and slowing down.

6

For s(t) = t³ − 12t on 0 ≤ t ≤ 3, find the total distance travelled.

7

A ball's height is h(t) = −4.9t² + 9.8t (m). It lands at t = 2 s. Find the total distance travelled and the average speed.

More practice: warm-up

1

v > 0 and a > 0. Speeding up or slowing down?

2

v < 0 and a > 0. Speeding up or slowing down?

3

An object moves 5 m right and then 3 m left. What is its displacement?

4

For the same motion, what is the total distance?

5

Can average speed ever be negative?

More practice: on your own

1

s(t) = −t² + 6t (m, s). When is the object slowing down?

2

s(t) = t³ − 6t² (m, s), t ≥ 0. Find all intervals of speeding up.

3

For s(t) = t³ − 6t² on 0 ≤ t ≤ 5, find the displacement and total distance.

4

s(t) = t⁴ − 8t² (m, s). Find the turning times for t ≥ 0 and the total distance on 0 ≤ t ≤ 3.

5

A particle has v(t) = 4 − t² (m/s) for t ≥ 0. Is it speeding up or slowing down at t = 1? At t = 3?

Challenge

1

s(t) = sin t (m), 0 ≤ t ≤ 2π. Find the total distance travelled.

2

At an instant when a(t) = 0 but v(t) ≠ 0, the sign rule gives no verdict. Explain why, and describe what is usually happening to the speed at that instant.

3

s(t) = t³ − 3t² − 9t + 5 (m, s), t ≥ 0. Find the total distance travelled in the first 4 seconds and the average speed.

Check your understanding

Quiz

1.

At a certain instant, v = −3 m/s and a = 2 m/s². What is happening?

2.

s(t) = t² − 4t (m, s). What is the total distance travelled from t = 0 to t = 5?

3.

Which statement about a ball thrown straight up (up positive) is true while it falls back down?

4.

An object moves along a line and returns to its starting point after 10 s, having travelled 40 m. What is its average velocity?

5.

s(t) = t³ − 3t (m, s), t ≥ 0. On which interval is the object slowing down?

6.

To find total distance travelled, you must first find…

For parents and teachers

Big idea (B2.1): an object speeds up when its velocity and acceleration have the same sign and slows down when they have opposite signs. Total distance travelled adds up every leg of the trip, while displacement only compares the start and the finish.

How to help: ask your teen to act out a motion with their hand along a table edge while you read the signs of v and a from their sign chart. If the hand speeds up when the chart says "slowing", something is wrong.

Common difficulties:

  • Thinking negative acceleration always means slowing down.
  • Adding signed displacements (which gives the displacement again) instead of their sizes when finding distance.
  • Missing a turning point, or counting a time when v = 0 but does not change sign.
  • Mixing up average velocity and average speed.