Motion Along a Line with Integrals

In motion with derivatives, you went forwards: position → velocity → acceleration. This page goes backwards. If you know how fast something is moving at every instant, integrating tells you where it ends up and how far it travelled. These two answers are often different, and AP questions love to test exactly that difference.

Key ideas

The chain, in both directions

For an object moving along a line with position $x(t)$:

$$x(t) \xrightarrow{\ \text{differentiate}\ } v(t) \xrightarrow{\ \text{differentiate}\ } a(t) \qquad\qquad a(t) \xrightarrow{\ \text{integrate}\ } v(t) \xrightarrow{\ \text{integrate}\ } x(t)$$

Integrating loses information (the $+ C$), so going backwards you need a starting value, called an initial condition, such as $x(0) = 5$.

Position from velocity

By the Fundamental Theorem of Calculus, the change in position from $0$ to $t$ is $\int_0^t v(u)\, du$. So

$$x(t) = x(0) + \int_0^t v(u)\, du$$

In words: where you are now = where you started + how much your position changed. The same idea works from any starting time: $x(b) = x(a) + \int_a^b v(t)\, dt$.

Likewise, velocity from acceleration: $v(t) = v(0) + \int_0^t a(u)\, du$.

Displacement vs total distance

	Formula	Meaning
Displacement on $[a, b]$	$\displaystyle\int_a^b v(t)\, dt$	net change in position, $x(b) - x(a)$; can be negative
Total distance on $[a, b]$	$\displaystyle\int_a^b \lvert v(t) \rvert\, dt$	every metre travelled counts as positive; never negative

Area above the $t$-axis means moving right (or up); area below means moving left (or down). Displacement lets the two cancel. Total distance adds them all up as positives.

Displacement is $\tfrac{4}{3} - \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{4}{3}$; total distance is $\tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4$.

Finding total distance by hand

To compute $\int_a^b \lvert v(t) \rvert\, dt$ without a calculator:

1. Solve $v(t) = 0$ to find where the object might turn around.
2. Split $[a, b]$ at those times.
3. Integrate $v$ on each piece and add the absolute values of the results.

Equivalently, find the position at $a$, at each turning point, and at $b$, and add up the distances between consecutive positions.

On a calculator-active question, you can simply enter $\int_a^b \lvert v(t) \rvert\, dt$ (use the abs function). Always write the integral down before the answer; AP graders look for it.

Worked examples

Example 1: Position from velocity

A particle moves along the $x$-axis with velocity $v(t) = 3t^2 - 2t$ (metres per second). At $t = 0$ it is at $x = 5$. Where is it at $t = 2$?

Solution.

$$x(2) = x(0) + \int_0^2 (3t^2 - 2t)\, dt = 5 + \Big[ t^3 - t^2 \Big]_0^2 = 5 + (8 - 4) = 9$$

The particle is at $x = 9$ m.

Example 2: Displacement vs total distance

A particle has velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \le t \le 4$. Find its displacement and its total distance travelled.

Solution. Displacement:

$$\int_0^4 (t^2 - 4t + 3)\, dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_0^4 = \frac{64}{3} - 32 + 12 = \frac{4}{3}$$

For distance, find where $v = 0$: $t^2 - 4t + 3 = (t - 1)(t - 3) = 0$, so $t = 1$ and $t = 3$. Let $F(t) = \dfrac{t^3}{3} - 2t^2 + 3t$. Then $F(0) = 0$, $F(1) = \dfrac{4}{3}$, $F(3) = 0$, $F(4) = \dfrac{4}{3}$, so

$$\int_0^1 v\, dt = \frac{4}{3}, \qquad \int_1^3 v\, dt = -\frac{4}{3}, \qquad \int_3^4 v\, dt = \frac{4}{3}$$

Total distance $= \dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4$ m. The displacement is only $\dfrac{4}{3}$ m because the particle backed up for $2$ seconds in the middle.

Example 3: From acceleration all the way to position

A particle has acceleration $a(t) = 6t - 6$, with $v(0) = -9$ and $x(0) = 2$. Find $x(t)$, and the total distance travelled from $t = 0$ to $t = 4$.

Solution. Velocity:

$$v(t) = -9 + \int_0^t (6u - 6)\, du = 3t^2 - 6t - 9 = 3(t - 3)(t + 1)$$

Position:

$$x(t) = 2 + \int_0^t (3u^2 - 6u - 9)\, du = t^3 - 3t^2 - 9t + 2$$

On $[0, 4]$, $v = 0$ only at $t = 3$ (the root $t = -1$ is outside the interval). Use positions:

$$x(0) = 2, \qquad x(3) = 27 - 27 - 27 + 2 = -25, \qquad x(4) = 64 - 48 - 36 + 2 = -18$$

The particle moves left from $2$ to $-25$ (that’s $27$ m), then right from $-25$ to $-18$ ($7$ m). Total distance $= 27 + 7 = 34$ m.

Example 4: Calculator active

A particle moves along a line with $v(t) = 2\sin(1.5t) + 0.5$ m/s for $0 \le t \le 4$, and $x(0) = 1$. (Calculator in radian mode.) Find (a) the position at $t = 4$, (b) the total distance travelled.

Solution. (a)

$$x(4) = 1 + \int_0^4 \big( 2\sin(1.5t) + 0.5 \big)\, dt \approx 1 + 2.053 = 3.053$$

(b)

$$\text{distance} = \int_0^4 \lvert 2\sin(1.5t) + 0.5 \rvert\, dt \approx 5.458 \text{ m}$$

Check that this is reasonable: $v$ changes sign once in $[0, 4]$, at $t \approx 2.263$, so the particle turns around. Distance should be more than the displacement of $2.053$ m, and it is.

Common mistakes

Forgetting the initial position. $\int_0^t v\, dt$ is the change in position, not the position. Add $x(0)$.

Using displacement when the question asks for distance. “How far did it travel?” or “total distance” means $\int \lvert v \rvert\, dt$. “Where is it?” or “displacement” means $\int v\, dt$ (plus a starting point, for position).

Taking the absolute value of the final answer. $\left\lvert \int v\, dt \right\rvert$ is not the same as $\int \lvert v \rvert\, dt$. In Example 2, the first gives $\tfrac{4}{3}$ and the second gives $4$.

Missing a turning point, or keeping one outside the interval. Solve $v(t) = 0$ and keep only the times inside $[a, b]$.

Calling acceleration’s sign “speeding up.” Speeding up needs $v$ and $a$ to have the same sign. That’s a derivative idea, but it often appears in the same AP question, so read each part carefully.

Calculator in degree mode. AP trig is always in radians. A wrong mode silently gives a wrong integral.

Practice

1. (Warm-up) A particle has velocity $v(t) = 4t + 1$ and $x(0) = 3$. Find $x(2)$.

Solution

$$x(2) = 3 + \int_0^2 (4t + 1)\, dt = 3 + \Big[ 2t^2 + t \Big]_0^2 = 3 + 10 = 13$$

2. (Warm-up) A particle has velocity $v(t) = t - 2$ for $0 \le t \le 4$. Find its displacement and its total distance travelled.

Solution

The graph of $v$ is a line through $(2, 0)$: a triangle below the axis on $[0, 2]$ and a triangle above on $[2, 4]$, each with area $\tfrac{1}{2}(2)(2) = 2$.

Displacement $= -2 + 2 = 0$. Total distance $= 2 + 2 = 4$.

The particle ends exactly where it started, even though it travelled $4$ units.

3. (Warm-up) An object has constant acceleration $a(t) = 6$ m/s² and $v(0) = 2$ m/s. Find $v(3)$.

Solution

$$v(3) = 2 + \int_0^3 6\, dt = 2 + 18 = 20 \text{ m/s}$$

4. (Core) A particle has velocity $v(t) = t^2 - 6t + 8$ for $0 \le t \le 5$. Find the displacement and the total distance travelled.

Solution

Let $F(t) = \dfrac{t^3}{3} - 3t^2 + 8t$. Displacement $= F(5) - F(0) = \dfrac{125}{3} - 75 + 40 = \dfrac{20}{3}$.

$v(t) = (t - 2)(t - 4)$ is zero at $t = 2$ and $t = 4$. With $F(0) = 0$, $F(2) = \dfrac{20}{3}$, $F(4) = \dfrac{16}{3}$, $F(5) = \dfrac{20}{3}$:

$$\int_0^2 v\, dt = \frac{20}{3}, \qquad \int_2^4 v\, dt = -\frac{4}{3}, \qquad \int_4^5 v\, dt = \frac{4}{3}$$

Total distance $= \dfrac{20}{3} + \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{28}{3} \approx 9.333$.

5. (Core) A ball is thrown straight up from a height of $2$ m with initial velocity $14.7$ m/s. Its acceleration is $a(t) = -9.8$ m/s². Find its maximum height.

Solution

$v(t) = 14.7 + \int_0^t (-9.8)\, du = 14.7 - 9.8t$, and

$$h(t) = 2 + \int_0^t (14.7 - 9.8u)\, du = 2 + 14.7t - 4.9t^2$$

The ball is highest when $v(t) = 0$: $t = \dfrac{14.7}{9.8} = 1.5$ s (velocity changes from positive to negative there).

$$h(1.5) = 2 + 22.05 - 11.025 = 13.025 \text{ m}$$

6. (Core) A particle has velocity $v(t) = \sin t$ for $0 \le t \le \dfrac{3\pi}{2}$ (radians). Find the displacement and the total distance travelled.

Solution

Displacement:

$$\int_0^{3\pi/2} \sin t\, dt = \Big[ -\cos t \Big]_0^{3\pi/2} = -0 + 1 = 1$$

$\sin t$ is positive on $(0, \pi)$ and negative on $\left(\pi, \tfrac{3\pi}{2}\right)$:

$$\int_0^{\pi} \sin t\, dt = 2, \qquad \int_{\pi}^{3\pi/2} \sin t\, dt = \Big[ -\cos t \Big]_{\pi}^{3\pi/2} = 0 - 1 = -1$$

Total distance $= 2 + 1 = 3$.

7. (Core) A particle moves along the $x$-axis with $v(t) = 2t - 6$ for $0 \le t \le 5$, and $x(0) = 4$.

• (a) Find $x(5)$.
• (b) When is the particle farthest to the left? Justify.
• (c) Find the total distance travelled.

Solution

$x(t) = 4 + \int_0^t (2u - 6)\, du = t^2 - 6t + 4$.

(a) $x(5) = 25 - 30 + 4 = -1$.

(b) $v(t) = 0$ at $t = 3$. Since $v$ changes from negative to positive at $t = 3$, the particle moves left and then right, so it is farthest left at $t = 3$, where $x(3) = 9 - 18 + 4 = -5$. (Compare the endpoints: $x(0) = 4$, $x(5) = -1$, both to the right of $-5$.)

(c) From $4$ to $-5$ is $9$ units; from $-5$ to $-1$ is $4$ units. Total distance $= 13$.

8. (Challenge) (Calculator active.) A particle moves along a line with $v(t) = \ln(t^2 + 1) - 1$ for $0 \le t \le 3$, and $x(0) = -2$.

• (a) Find $x(3)$.
• (b) At what time does the particle change direction?
• (c) Find the total distance travelled.

Solution

(a)

$$x(3) = -2 + \int_0^3 \big( \ln(t^2 + 1) - 1 \big)\, dt \approx -2 + 0.406 = -1.594$$

(b) $v(t) = 0$ when $\ln(t^2 + 1) = 1$, so $t^2 + 1 = e$ and $t = \sqrt{e - 1} \approx 1.311$. $v$ changes from negative to positive there, so the particle changes direction at $t \approx 1.311$.

(c)

$$\int_0^3 \lvert \ln(t^2 + 1) - 1 \rvert\, dt \approx 1.973$$

Check: the two pieces are about $-0.783$ (on $[0, 1.311]$) and $1.189$ (on $[1.311, 3]$); their sum is the displacement $0.406$, and the sum of their sizes is the distance $1.973$ (up to rounding).

9. (Challenge) A particle has velocity $v(t) = t^2 - kt$ for $0 \le t \le 3$, where $k$ is a constant. Its displacement over $[0, 3]$ is $0$. Find $k$, then find the total distance travelled.

Solution

$$\int_0^3 (t^2 - kt)\, dt = 9 - \frac{9k}{2} = 0 \quad\Rightarrow\quad k = 2$$

Now $v(t) = t^2 - 2t = t(t - 2)$, negative on $(0, 2)$ and positive on $(2, 3)$. With $F(t) = \dfrac{t^3}{3} - t^2$:

$$\int_0^2 v\, dt = \frac{8}{3} - 4 = -\frac{4}{3}, \qquad \int_2^3 v\, dt = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}$$

Total distance $= \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{8}{3}$.