Motion with Integrals — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

A runner jogs 3 km out along a trail and 3 km back. Her watch says she ran 6 km; her displacement is zero, because she ended where she started. Both numbers come from the same velocity graph. Integrate velocity and you get the change in position; integrate speed and you get the distance covered.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A particle has velocity $v(t)$ . Its speed is

$|v(t)|$
$v'(t)$
$-v(t)$
$v(t)^2$

Show the answer

Speed is the size of the velocity, without its direction.

Correct: $|v(t)|$ :
$v'(t)$ :
$-v(t)$ :
$v(t)^2$ :

2. If $F' = f$ is continuous, then $\int_a^b f(t)\,dt$ $=$

$F(b) - F(a)$
$F(a) - F(b)$
$f(b) - f(a)$
$F(b)$

Show the answer

The integral of a rate of change is the net change.

Correct: $F(b) - F(a)$ :
$F(a) - F(b)$ :
$f(b) - f(a)$ :
$F(b)$ :

Part 4 · See it

See it first

The velocity v(t) = t² − 4t + 3 m/s for 0 ≤ t ≤ 4 s. It is positive on (0, 1) and (3, 4), where the region between the graph and the t-axis is shaded crimson with area 4/3 each, and negative on (1, 3), where the region below the axis is shaded blue with area 4/3. Displacement adds signed areas: 4/3 m. Total distance adds the areas: 4 m. — $v(t) = t^2 - 4t + 3$ on $[0, 4]$ . Signed areas give the displacement, $\frac{4}{3}$ m; the areas themselves give the total distance, 4 m. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$v(t) = x'(t)$ $\int_a^b v(t)\,dt$ $= x(b) - x(a)$ , the displacement
Area below the t-axis counts as negative in that integralmotion to the left cancels motion to the right
Speed is $|v(t)|$ $\int_a^b |v(t)|\,dt$ counts every stretch as positive: the total distance
The position at one time $x(a)$ is known $x(t)$ $= x(a)$ $+ \int_a^t v(s)\,ds$ gives the position at any time

Part 6 · Key ideas

Key ideas

Displacement on $[a, b]$ : $\int_a^b v(t)\,dt$ $= x(b) - x(a)$ . Its sign gives the direction of the net change.
Total distance traveled: $\int_a^b |v(t)|\,dt$ . Without a calculator, split at the zeros of $v$ and add the sizes of the pieces.
Position: $x(t)$ $= x(t_0)$ $+ \int_{t_0}^{t} v(s)\,ds$ . The initial position is not optional.
Velocity from acceleration works the same way: $v(t)$ $= v(t_0)$ $+ \int_{t_0}^{t} a(s)\,ds$ .
Units: integrating meters per second over seconds gives meters.

Part 7 · Misconception

A common mistake

The wrong idea: The total distance on $[0, 4]$ for $v(t) = t^2 - 4t + 3$ is $\int_0^4 v(t)\,dt$ $= \frac{4}{3}$ m.

What actually happens: That is the displacement. $v < 0$ on $(1, 3)$ , so the particle moves left and that stretch subtracts. Total distance: $\int_0^4 |v(t)|\,dt$ $= \frac{4}{3}$ $+ \frac{4}{3}$ $+ \frac{4}{3}$ $= 4$ m.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The velocity $v(t)$ of a particle

1. Find the displacement of the particle from $t = 0$ to $t = 10$ seconds. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\int_0^{10} v(t)\,dt$ $= 4$ $+ 12$ $+ 4 - 1 - 4$ $= 15$ meters.

Answer: 15 meters

2. Find the total distance the particle travels from $t = 0$ to $t = 10$ seconds. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\int_0^{10} |v(t)|\,dt$ $= 4$ $+ 12$ $+ 4$ $+ 1$ $+ 4$ $= 25$ meters.

Answer: 25 meters

3. The particle is at $x = -3$ meters at $t = 0$ . Find its position at $t = 10$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine), in meters.

Show the answer

$x(10)$ $= x(0)$ $+ \int_0^{10} v(t)\,dt$ $= -3$ $+ 15$ $= 12$ meters.

Answer: 12 meters

4. Which expression gives the total distance a particle with velocity $v(t)$ travels from $t = 1$ to $t = 4$ ?

$\left|\int_1^4 v(t)\,dt\right|$
$\int_1^4 v(t)\,dt$
$\int_1^4 v'(t)\,dt$
$\int_1^4 |v(t)|\,dt$

Show the answer

Total distance $= \int_1^4 |v(t)|\,dt$ ; the absolute value goes inside the integral.

$\left|\int_1^4 v(t)\,dt\right|$ : That is the size of the displacement; back-and-forth motion cancels inside.
$\int_1^4 v(t)\,dt$ : That is the displacement.
$\int_1^4 v'(t)\,dt$ : That is $v(4) - v(1)$ , a change in velocity.
Correct: $\int_1^4 |v(t)|\,dt$ : The integral of speed counts every stretch as positive.

5. A particle has velocity $v(t) = 3t^2 - 2$ and position $x(1) = 4$ . Find $x(t)$ .

Type an expression for $x(t) =$ .

Show the answer

$x(t)$ $= x(1)$ $+ \int_1^t (3s^2 - 2)\,ds$ $= 4$ $+ (t^3 - 2t) - (1 - 2)$ $= t^3 - 2t$ $+ 5$ .

Answer: $x(t) =$ $t^{3} - 2t + 5$

6. A particle moves with $v(t) = t^2 - 4t + 3$ m/s. Find the total distance it travels on $0 \le t \le 4$ , without a calculator.

No calculator. Type the exact value (a fraction, π, e or a root is fine), in meters.

Show the answer

$v = (t - 1)(t - 3)$ changes sign at 1 and 3. $\int_0^1 v = \frac{4}{3}$ , $\int_1^3 v$ $= -\frac{4}{3}$ , $\int_3^4 v = \frac{4}{3}$ . Total distance $= 4$ m.

Answer: 4 meters

Part 9 · Summary

Summary

Integrate velocity for displacement,

x(b) - x(a)

; integrate speed

|v|

for total distance. To find a position, start from a known position and add the integral of velocity. Split at the zeros of

v

when working by hand, and always state units.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The velocity v(t)v(t) of a particle

Summary

What comes next

Connections

The velocity $v(t)$ of a particle