Math Core

Motion along a line is the most tested context in AP Calculus. It appears on nearly every free-response section, and it rewards one habit above all: keep position, velocity, acceleration and speed separate, and know exactly what the sign of each one tells you.

Three functions, two derivatives

Picture a particle moving back and forth along the $x$ -axis. Its position at time $t$ is $x(t)$ (some books write $s(t)$ ). The rate of change of position is the velocity, and the rate of change of velocity is the acceleration:

v(t) = x'(t), \qquad a(t) = v'(t) = x''(t).

If position is measured in meters and time in seconds, then velocity is in meters per second and acceleration is in meters per second per second (m/s²).

Definition

Velocity, speed and acceleration

For a particle with position $x(t)$ on a line:

Velocity $v(t) = x'(t)$ gives both rate and direction. $v(t) > 0$ means the particle moves right (or up); $v(t) < 0$ means it moves left (or down).
Speed is $|v(t)|$ , the size of the velocity with no direction. Speed is never negative.
Acceleration $a(t) = v'(t)$ is the rate of change of velocity, not of speed.

What the signs tell you

Most motion questions are really sign questions.

At rest: $v(t) = 0$ .
Changes direction: $v(t)$ changes sign. A zero of $v$ is not enough by itself; the sign has to switch from positive to negative or the reverse.
Moving right / left: $v(t) > 0$ / $v(t) < 0$ .

The tricky one is whether the particle is speeding up or slowing down. You might guess that positive acceleration means speeding up, but that's only true when the particle is moving in the positive direction. Think of a car moving backward while you press the forward pedal: the acceleration is positive, yet the car slows down.

Speeding up or slowing down

If $v(t)$ and $a(t)$ have the same sign, the speed is increasing: the particle is speeding up.
If $v(t)$ and $a(t)$ have opposite signs, the speed is decreasing: the particle is slowing down.

Common mistake

" $a(t) > 0$ , so the particle is speeding up" earns no credit on the AP exam. You must compare the signs of velocity and acceleration, and your written justification should mention both, for example: " $v(2.5) < 0$ and $a(2.5) > 0$ , so the particle is slowing down."

A sign chart does the work

For a polynomial position function, the standard approach is:

Differentiate to get $v(t)$ and $a(t)$ .
Find the zeros of each.
Put both on one sign chart and read off direction, and speeding up versus slowing down, on each interval.

Position x(t) = t³ − 6t² + 9t. The particle turns around at t = 1 and t = 3, where the graph has horizontal tangents.Open in grapher →

Worked example: At rest and moving left

A particle moves along the $x$ -axis with position $x(t) = t^3 - 6t^2 + 9t$ for $t \ge 0$ . When is the particle at rest, and when is it moving left?

Solution. $v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ . The particle is at rest when $v(t) = 0$ : at $t = 1$ and $t = 3$ .

Test the sign of $v$ on each interval: $v(0.5) > 0$ , $v(2) = -3 < 0$ , $v(4) > 0$ . The particle moves left on $1 < t < 3$ and right on $0 \le t < 1$ and $t > 3$ . It changes direction at both $t = 1$ and $t = 3$ .

Worked example: Speeding up or slowing down

For the same particle, is it speeding up or slowing down at $t = 2.5$ ? On which intervals is it speeding up?

Solution. $a(t) = 6t - 12$ , which is zero at $t = 2$ .

At $t = 2.5$ : $v(2.5) = 3(1.5)(-0.5) = -2.25 < 0$ and $a(2.5) = 3 > 0$ . The signs differ, so the particle is slowing down.

Combining the sign changes of $v$ (at $t = 1$ and $t = 3$ ) and of $a$ (at $t = 2$ ):

interval	$0 < t < 1$	$1 < t < 2$	$2 < t < 3$	$t > 3$
sign of $v$	$+$	$-$	$-$	$+$
sign of $a$	$-$	$-$	$+$	$+$
motion	slowing	speeding up	slowing	speeding up

The particle speeds up on $1 < t < 2$ and $t > 3$ .

Worked example: Trigonometric motion

An object on a spring has position $x(t) = 4\sin\!\left(\dfrac{\pi t}{2}\right)$ centimeters at time $t$ seconds. Find its velocity and acceleration at $t = 1$ .

Solution. By the chain rule,

v(t) = 2\pi\cos\!\left(\frac{\pi t}{2}\right), \qquad a(t) = -\pi^2\sin\!\left(\frac{\pi t}{2}\right).

At $t = 1$ : $v(1) = 2\pi\cos(\pi/2) = 0$ cm/s and $a(1) = -\pi^2 \approx -9.87$ cm/s². The object is momentarily at rest at its highest point, and the negative acceleration pulls it back down.

Vertical motion

When an object is thrown straight up, its height $h(t)$ plays the role of position, and "up" is the positive direction. Near Earth's surface, height in feet is often modeled by $h(t) = -16t^2 + v_0 t + h_0$ , where $v_0$ is the initial velocity and $h_0$ the starting height.

Worked example: A ball thrown upward

A ball is thrown from the top of an 80-foot building with height $h(t) = -16t^2 + 64t + 80$ feet. Find its maximum height and its velocity when it hits the ground.

Solution. $v(t) = -32t + 64$ , which is zero at $t = 2$ . The ball rises until $t = 2$ , so the maximum height is $h(2) = -64 + 128 + 80 = 144$ feet.

It hits the ground when $h(t) = 0$ : $-16(t^2 - 4t - 5) = 0$ , so $(t - 5)(t + 1) = 0$ and $t = 5$ (reject $t = -1$ ). Its velocity then is $v(5) = -160 + 64 = -96$ feet per second. The negative sign says the ball is moving down; its speed at impact is 96 ft/s.

Tip

At the highest point of a vertical path, velocity is 0 but acceleration is not. Here $a(t) = -32$ ft/s² the whole time, including at the top.

Practice

Practice 1

A particle moves along a line with position $x(t) = t^2 - 8t + 3$ meters, $t$ in seconds. Find its velocity at $t = 5$ , in meters per second.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A particle has position $x(t) = t^3 - 3t^2 + 5$ . Find its acceleration at $t = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A particle moves with position $x(t) = 2t^3 - 15t^2 + 36t$ for $t \ge 0$ . At what times is the particle at rest? List all of them.

Separate answers with commas, e.g. 2, -5

Practice 4

A particle moves along the $x$ -axis with position $x(t) = t^3 - 9t^2 + 24t$ for $t \ge 0$ . On what open interval is the particle moving to the left? Write your answer as an inequality in $t$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 5

A particle has velocity $v(t) = t^2 - 5t + 4$ . Which statement about the particle at $t = 2$ is true?

Practice 6

A stone is thrown upward from a 64-foot cliff. Its height is $h(t) = -16t^2 + 48t + 64$ feet after $t$ seconds. Find its velocity, in feet per second, at the moment it hits the ground.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

(Calculator, part a.) A particle moves along the $x$ -axis with velocity $v(t) = \sin(t^2)$ for $0 \le t \le 3$ . Find the acceleration of the particle at $t = 2$ . Round to three decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

(Part b.) For the particle in part (a), is the speed of the particle increasing or decreasing at $t = 2$ ?