Straight-Line Motion with Derivatives

A “particle moving along a line” shows up on almost every calculus test (and on almost every AP Calculus exam). If you know its position as a function of time, derivatives tell you everything else: how fast it’s going, which way it’s moving, and whether it’s speeding up or slowing down. This page builds the tools and the justification language you need.

Key ideas

Position, velocity, acceleration

A particle moves along a horizontal line (the $x$-axis). Its position at time $t$ is $x(t)$ (some books use $s(t)$).

Quantity	Formula	Typical units
Position	$x(t)$	m
Velocity	$v(t) = x'(t)$	m/s
Acceleration	$a(t) = v'(t) = x''(t)$	m/s²

Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

Direction and speed

• $v(t) \gt 0$: the particle is moving right (in the positive direction).
• $v(t) \lt 0$: it is moving left.
• $v(t) = 0$: it is at rest at that instant.
• Speed is the size of the velocity: $\text{speed} = |v(t)|$. Speed is never negative.

The particle changes direction at a time when $v(t)$ changes sign. Being at rest isn’t enough: $v$ has to go from positive to negative or the other way.

Speeding up or slowing down

Compare the signs of $v$ and $a$:

• Same sign (both positive or both negative): the particle is speeding up (speed increasing).
• Opposite signs: it is slowing down.

This works because acceleration in the same direction as the motion pushes the particle faster, and acceleration against the motion slows it. A negative acceleration doesn’t always mean slowing down: if the particle is moving left ($v \lt 0$) and $a \lt 0$, it’s speeding up.

A sign chart does most of the work

Find where $v(t) = 0$ and where $a(t) = 0$. These times split the time interval into pieces. Test the sign of $v$ and of $a$ in each piece, then read off direction and speeding up or slowing down.

On a test, justify with signs: “At $t = 2.5$, $v(2.5) \lt 0$ and $a(2.5) \lt 0$, so the particle is speeding up.”

Worked examples

Example 1: Values at one instant

A particle’s position in metres is $x(t) = t^2 - 4t + 3$ for $t \ge 0$ seconds. At $t = 1$, find the position, velocity, speed, and acceleration, and describe the motion.

Solution.

$$v(t) = 2t - 4, \qquad a(t) = 2$$

At $t = 1$: $x(1) = 1 - 4 + 3 = 0$, $v(1) = -2$, speed $= |-2| = 2$, and $a(1) = 2$.

The particle is at the origin, moving left at $2$ m/s. Since $v(1) \lt 0$ and $a(1) \gt 0$ have opposite signs, it is slowing down.

Example 2: A full analysis

A particle moves with position $x(t) = t^3 - 9t^2 + 24t$ metres for $0 \le t \le 5$ seconds.

(a) When is the particle at rest? When does it change direction?

(b) On which intervals is it speeding up? Slowing down?

(c) Find the total distance travelled from $t = 0$ to $t = 5$.

Solution.

$$v(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4), \qquad a(t) = 6t - 18 = 6(t - 3)$$

(a) $v(t) = 0$ at $t = 2$ and $t = 4$, so the particle is at rest at those times. $v$ changes sign at both (positive, then negative, then positive), so the particle changes direction at $t = 2$ and at $t = 4$.

(b) Make a sign chart using $t = 2, 3, 4$:

Interval	$(0, 2)$	$(2, 3)$	$(3, 4)$	$(4, 5)$
Sign of $v$	$+$	$-$	$-$	$+$
Sign of $a$	$-$	$-$	$+$	$+$
Direction	right	left	left	right
Speed	slowing down	speeding up	slowing down	speeding up

It is speeding up on $(2, 3)$ and $(4, 5)$, and slowing down on $(0, 2)$ and $(3, 4)$.

Where the graph moves toward the $t$-axis, speed decreases; where it moves away from the axis, speed increases.

(c) The particle turns around at $t = 2$ and $t = 4$, so find the position at each turning point and the endpoints:

$$x(0) = 0, \quad x(2) = 20, \quad x(4) = 16, \quad x(5) = 20$$

It goes right $20$ m, back left $4$ m, then right $4$ m. Total distance $= 20 + 4 + 4 = 28$ m. (Its final position is $20$ m from the start, but it travelled $28$ m.)

Example 3: Trig motion (calculator active)

A particle’s position is $x(t) = 2\sin t$ for $t \ge 0$, with $t$ in seconds and the angle in radians. Is the particle speeding up or slowing down at $t = 2$?

Solution.

$$v(t) = 2\cos t, \qquad a(t) = -2\sin t$$ $$v(2) = 2\cos 2 \approx -0.832, \qquad a(2) = -2\sin 2 \approx -1.819$$

Both are negative, so the particle is speeding up at $t = 2$. (Make sure your calculator is in radian mode: $\cos 2$ means $2$ radians, about $115^\circ$.)

Common mistakes

Saying the particle changes direction whenever $v = 0$. It must change sign. For $v(t) = (t - 1)^2$, the particle stops at $t = 1$ but keeps moving right.

Thinking negative acceleration means slowing down. Compare the signs of $v$ and $a$. Negative velocity with negative acceleration means speeding up.

Mixing up speed and velocity. Speed is $|v(t)|$. A velocity of $-5$ m/s is a speed of $5$ m/s. The speed is never negative.

Using $x(5) - x(0)$ as the total distance. That’s the displacement. If the particle turns around, add the distance of each leg separately.

Using the sign of $x(t)$ to decide direction. A particle can be to the left of the origin ($x \lt 0$) while moving right ($v \gt 0$). Direction comes from $v$, not $x$.

Writing a vague justification. Markers (including AP graders) want the reason: ”$v(3) \lt 0$ and $a(3) \gt 0$, so the particle is slowing down”, not just “slowing down”.

Practice

1. (Warm-up) A particle has position $x(t) = t^3 - 3t$. Find $v(2)$ and $a(2)$.

Solution

$v(t) = 3t^2 - 3$ and $a(t) = 6t$.

$v(2) = 12 - 3 = 9$ and $a(2) = 12$.

2. (Warm-up) At $t = 3$ s, a particle’s velocity is $-5$ m/s. What is its speed, and which way is it moving?

Solution

Its speed is $|-5| = 5$ m/s, and it is moving left (the negative direction).

3. (Warm-up) At some instant, $v = -4$ m/s and $a = 2$ m/s². Is the particle speeding up or slowing down?

Solution

The signs are opposite, so it is slowing down.

4. (Core) A particle has position $x(t) = t^2 - 6t + 5$ metres for $0 \le t \le 5$ seconds. When does it change direction? Find the total distance travelled.

Solution

$v(t) = 2t - 6$, which is negative for $t \lt 3$ and positive for $t \gt 3$. The particle changes direction at $t = 3$.

$x(0) = 5$, $x(3) = -4$, $x(5) = 0$.

It moves left $9$ m (from $5$ to $-4$), then right $4$ m (from $-4$ to $0$). Total distance $= 9 + 4 = 13$ m.

5. (Core) A particle has position $x(t) = 2t^3 - 15t^2 + 36t$ for $t \ge 0$. On which intervals is it speeding up?

Solution

$$v(t) = 6t^2 - 30t + 36 = 6(t - 2)(t - 3), \qquad a(t) = 12t - 30 = 6(2t - 5)$$

$v = 0$ at $t = 2, 3$ and $a = 0$ at $t = 2.5$.

Interval	$(0, 2)$	$(2, 2.5)$	$(2.5, 3)$	$t \gt 3$
$v$	$+$	$-$	$-$	$+$
$a$	$-$	$-$	$+$	$+$

The signs match on $(2, 2.5)$ and for $t \gt 3$, so it is speeding up on those intervals.

6. (Core) A particle has position $x(t) = te^{-t}$ for $t \ge 0$.

• (a) When is the particle farthest to the right?
• (b) Is it speeding up or slowing down at $t = 1.5$?

Solution

By the product rule:

$$v(t) = e^{-t} - te^{-t} = (1 - t)e^{-t}, \qquad a(t) = -e^{-t} - (1 - t)e^{-t} = (t - 2)e^{-t}$$

(a) $v \gt 0$ for $t \lt 1$ and $v \lt 0$ for $t \gt 1$, so the particle moves right, then left. It is farthest right at $t = 1$, where $x(1) = e^{-1} \approx 0.368$.

(b) $v(1.5) = -0.5e^{-1.5} \lt 0$ and $a(1.5) = -0.5e^{-1.5} \lt 0$. Same sign, so it is speeding up.

7. (Core) A ball is thrown upward. Its height in metres is $h(t) = -4.9t^2 + 14.7t + 2$, $t$ in seconds.

• (a) Find the maximum height.
• (b) Find the ball’s velocity when it hits the ground (calculator active; 3 decimal places).

Solution

(a) $v(t) = -9.8t + 14.7 = 0$ at $t = 1.5$. Then

$$h(1.5) = -4.9(2.25) + 14.7(1.5) + 2 = -11.025 + 22.05 + 2 = 13.025$$

The maximum height is $13.025$ m.

(b) Solve $h(t) = 0$ with the quadratic formula (or a calculator). The positive root is $t \approx 3.130$ s. Then

$$v(3.130) \approx -9.8(3.13039) + 14.7 \approx -15.978$$

The ball hits the ground with velocity about $-15.978$ m/s (moving down at about $15.978$ m/s).

8. (Core) A particle has position $x(t) = \cos(2t)$ for $0 \le t \le \pi$ (radians). When is it at rest, and when does it change direction? Find its speed at $t = \dfrac{\pi}{3}$.

Solution

$v(t) = -2\sin(2t)$. On $[0, \pi]$, $\sin(2t) = 0$ when $2t = 0, \pi, 2\pi$, so the particle is at rest at $t = 0$, $\dfrac{\pi}{2}$, and $\pi$.

$v \lt 0$ on $\left(0, \dfrac{\pi}{2}\right)$ and $v \gt 0$ on $\left(\dfrac{\pi}{2}, \pi\right)$, so it changes direction at $t = \dfrac{\pi}{2}$ only. (At the endpoints there’s no sign change inside the interval.)

$$\left|v\left(\frac{\pi}{3}\right)\right| = \left|-2\sin\frac{2\pi}{3}\right| = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$

9. (Challenge) A particle has position $x(t) = t^3 - 6t^2 + 9t + 1$ for $0 \le t \le 4$. Find its maximum speed on this interval.

Solution

$v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ and $a(t) = 6t - 12$.

The speed is $|v(t)|$. The largest value of $|v|$ happens where $v$ has its largest or smallest value, so check the endpoints and where $a = v' = 0$ (at $t = 2$):

$$v(0) = 9, \qquad v(2) = -3, \qquad v(4) = 9$$

Speeds: $9$, $3$, $9$. The maximum speed is $9$, at $t = 0$ and $t = 4$.

10. (Challenge) Two particles move on the same line with positions $x_1(t) = t^2 - 4t$ and $x_2(t) = -t^2 + 2t$ for $t \ge 0$. During what time interval are they moving in the same direction?

Solution

$v_1(t) = 2t - 4$ and $v_2(t) = -2t + 2$. They move in the same direction when the velocities have the same sign, so $v_1 v_2 \gt 0$:

$$(2t - 4)(-2t + 2) \gt 0 \quad\Leftrightarrow\quad -4(t - 2)(t - 1) \gt 0 \quad\Leftrightarrow\quad (t - 1)(t - 2) \lt 0$$

So $1 \lt t \lt 2$. (On that interval, both velocities are negative: both particles move left.)