Straight-Line Motion | Math Coach Amy

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Key Ideas Easy Medium Hard

\(v = s',\quad a = v'\)

speeding up when v and a have the same sign

What to Remember

Position, velocity, acceleration

\(v(t) = s'(t)\) and \(a(t) = v'(t) = s''(t)\).

On these pages, positive velocity means motion in the positive direction.

At rest, then the direction

The particle is at rest when \(v(t) = 0\).

It moves in the positive direction when \(v(t) > 0\) and in the negative direction when \(v(t) < 0\).

Speeding up uses both signs

The particle speeds up when \(v(t)\) and \(a(t)\) have the same sign. It slows down when the signs are opposite.

Distance from a position formula

Displacement from \(t = a\) to \(t = b\) is \(s(b) - s(a)\).

If you also know the times when the particle turns around, the total distance is the sum of the absolute changes in position between those times.

Level 1

Easy

Find velocity and acceleration, then read the signs.

Example

A particle moves on a line with \(s(t) = t^2 - 4t\), \(t \ge 0\), in meters, with \(t\) in seconds. When is the particle at rest?

\(v(t) = s'(t) = 2t - 4\)

\(2t - 4 = 0\)

\(t = 2\)

Answer: \(t = 2\) seconds

For \(s(t) = t^2 - 4t\), \(t \ge 0\), on which intervals is the particle moving in the positive direction?

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\(v(t) = 2t - 4 = 2(t - 2)\)

\(v(t) > 0\) when \(t > 2\).

Answer: \((2, \infty)\)

The velocity of a particle is \(v(t) = t^2 - 9\) for \(t \ge 0\). When is the particle at rest?

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\(t^2 - 9 = 0\)

\(t = 3\), since \(t \ge 0\).

Answer: \(t = 3\)

A particle has \(v(t) = 4\) for all \(t\). Describe its acceleration and whether it is speeding up.

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\(a(t) = 0\)

Velocity is constant and positive.

The particle is moving in the positive direction and is neither speeding up nor slowing down.

Answer: a = 0; neither speeding up nor slowing down

At a certain time, \(v > 0\) and \(a < 0\). Is the particle speeding up or slowing down?

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The signs of velocity and acceleration are opposite.

The particle is slowing down.

Answer: slowing down

For \(s(t) = 6t - t^2\), \(t \ge 0\), find \(a(t)\) and the time when the particle is at rest.

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\(v(t) = 6 - 2t\)

\(a(t) = -2\)

\(6 - 2t = 0\)

\(t = 3\)

Answer: \(a(t) = -2\); at rest at \(t = 3\)

Level 2

Medium

Split the timeline at the zeros of velocity and the zero of acceleration.

Example

A particle has position \(s(t) = t^3 - 6t^2 + 9t\) for \(t \ge 0\). Find the times when it is at rest.

\(v(t) = 3t^2 - 12t + 9\)

\(v(t) = 3(t - 1)(t - 3)\)

\(t = 1\) and \(t = 3\)

Answer: \(t = 1\) and \(t = 3\)

For \(s(t) = t^3 - 6t^2 + 9t\), \(t \ge 0\), on which intervals is the particle moving in the negative direction?

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\(v(t) = 3(t - 1)(t - 3)\)

Test \(t = 2\): \(v(2) = -3 < 0\).

Test \(t = 0\): \(v(0) = 9 > 0\).

Test \(t = 4\): \(v(4) = 9 > 0\).

Negative velocity on \((1, 3)\).

Answer: \((1, 3)\)

For the same particle, find the intervals where it is speeding up.

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\(a(t) = 6t - 12 = 6(t - 2)\)

\(a(t) < 0\) on \([0, 2)\) and \(a(t) > 0\) on \((2, \infty)\).

Same signs on \((1, 2)\): both negative.

Same signs on \((3, \infty)\): both positive.

Answer: \((1, 2)\) and \((3, \infty)\)

A particle has position \(s(t) = t^3 - 3t^2\) for \(t \ge 0\). Find \(v(t)\) and the times when the particle is at rest.

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\(v(t) = 3t^2 - 6t\)

\(v(t) = 3t(t - 2)\)

\(t = 0\) and \(t = 2\)

Answer: \(v(t) = 3t(t - 2)\); at rest at 0 and 2

For \(s(t) = t^3 - 3t^2\), \(t \ge 0\), on which intervals is the particle speeding up?

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\(v(t) = 3t(t - 2)\)

\(v(t) < 0\) on \((0, 2)\) and \(v(t) > 0\) on \((2, \infty)\).

\(a(t) = 6t - 6 = 6(t - 1)\)

Both negative on \((0, 1)\).

Both positive on \((2, \infty)\).

Answer: \((0, 1)\) and \((2, \infty)\)

For \(s(t) = 6t - t^2\), is the particle speeding up or slowing down on \((0, 3)\)?

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\(v(t) = 6 - 2t > 0\) on \((0, 3)\).

\(a(t) = -2 < 0\)

The signs are opposite, so the particle is slowing down.

Answer: slowing down

Level 3

Hard

A velocity formula, a velocity graph, and distance versus displacement.

Example

The velocity of a particle, for \(t \ge 0\), is \(v(t) = (t - 1)(t - 4)\). On which intervals is the particle speeding up?

\(v(t) = (t - 1)(t - 4)\)

Positive on \([0, 1)\) and \((4, \infty)\). Negative on \((1, 4)\).

\(a(t) = 2t - 5\)

\(a(t) < 0\) on \([0, \dfrac{5}{2})\) and \(a(t) > 0\) on \((\dfrac{5}{2}, \infty)\).

Same signs on \((1, \dfrac{5}{2})\) and on \((4, \infty)\).

Answer: \((1, \dfrac{5}{2})\) and \((4, \infty)\)

Use the graph of \(v\). The first segment has slope \(-2\) and the second has slope 2. On which intervals is the particle slowing down?

Graph of v

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Velocity is positive on \((0, 2)\) and \((4, 6)\).

Velocity is negative on \((2, 4)\).

Acceleration is \(-2\) on \((0, 3)\) and 2 on \((3, 6)\).

Opposite signs on \((0, 2)\) and on \((3, 4)\).

Answer: \((0, 2)\) and \((3, 4)\)

For \(s(t) = t^3 - 6t^2 + 9t\), find the displacement from \(t = 0\) to \(t = 3\) and the total distance traveled on that interval.

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\(s(0) = 0\)

\(s(1) = 1 - 6 + 9 = 4\)

\(s(3) = 27 - 54 + 27 = 0\)

The particle is at rest at \(t = 1\) and \(t = 3\), and velocity changes sign at \(t = 1\).

Displacement: \(s(3) - s(0) = 0\)

Distance: \(|4 - 0| + |0 - 4| = 8\)

Answer: displacement 0 m; distance 8 m

For \(s(t) = t^3 - 3t^2\), find \(s(2)\) and say whether the particle is moving in the positive or negative direction just before \(t = 2\).

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\(s(2) = 8 - 12 = -4\)

\(v(t) = 3t(t - 2)\)

On \((0, 2)\), \(v(t) < 0\).

Just before \(t = 2\), the particle is moving in the negative direction, and \(s(2) = -4\).

Answer: \(s(2) = -4\); negative direction

The velocity graph is shown. At \(t = 3\), which statement is true?

Graph of v

A. The particle is at rest.

B. Velocity is negative and acceleration is not defined, because the graph has a corner.

C. The particle is speeding up, because the graph is below the axis.

D. Acceleration is 0.

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The closed point at \(t = 3\) is \((3, -2)\), so \(v(3) = -2\).

The slope on the left is \(-2\) and the slope on the right is 2.

Those slopes do not match, so \(a(3)\) does not exist.

Answer: B. \(v(3) = -2\) and \(a(3)\) does not exist

A particle has \(v(t) = t^2 - 5t + 4\) and \(a(t) = 2t - 5\). At \(t = 4\), is the particle at rest, and is it about to move in the positive direction?

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\(v(4) = 16 - 20 + 4 = 0\)

The particle is at rest at \(t = 4\).

For \(t > 4\), both factors \((t - 1)\) and \((t - 4)\) are positive, so \(v(t) > 0\).

Just after \(t = 4\), the particle moves in the positive direction.

Answer: at rest, then positive direction

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