4.8 Applications of Definite Integrals

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In this section we use definite integrals to study rectilinear motion.

1 Definite Integrals and Rectilinear Motion

Suppose that an object is moving along a straight path. The position, velocity, and acceleration functions governing the motion of the object can be derived from each other in the following ways.
If we start with the position function, $s(t)$ , we can compute velocity and acceleration by differentiating: $v(t) = s'(t)\quad \text {and} \quad a(t) = v'(t)$ On the other hand, if we start with the acceleration function, $a(t)$ , we can compute velocity and position by integrating: $v(t) = \int a(t) \ dt \quad \text {and} \quad s(t) = \int v(t) \ dt$ These indefinite integrals will contain a constant of integration, $C$ , which can be determined by using the initial conditions of the problem.

In the following proposition, we see that definite integrals that can be used to compute the displacement and distance traveled by an object moving along a straight path.

Suppose that an object moving along a straight path has instantaneous velocity $v(t)$ , then

(a): the displacement of the object over the interval $[a,b]$ is given by $s(b) - s(a) =\int _a^b v(t) \ dt$ and
(b): the distance traveled by the object over the interval $[a,b]$ is given by $\int _a^b |v(t)| \ dt.$

example 1 An object is launched vertically from with an initial velocity of $48 ft/sec$ . Assume that the acceleration of the object due to gravity is $-32 ft/sec^2$ .

(a): Find the displacement of the object from time $t = 0$ to time $t = 2$ .
(b): Find the total distance traveled from time $t = 0$ to time $t = 2$ .

Part (a): We fist find the velocity function: $v(t) = \int a(t) \ dt = \int -32 \ dt = -32t + C$ To find the value of the constant $C$ , we use the fact that the initial velocity was given as $48 ft/sec$ : $v(0) = -32(0) + C = 48,$ which implies that $C = 48$ . Thus, $v(t) = -32t + 48 \; \mbox { or } \; 48 - 32t.$ The displacement from $t= 0$ to $t = 2$ , is given by \begin{align*} s(2) - s(0) & = \int _0^2 v(t) \ dt \\ & = \int _0^2 (48 - 32t) \ dt \\ & = (48t - 16t^2)\bigg |_0^2 \\ & = 32 \text{ feet} \end{align*}

Part (b): To find the total distance traveled from $t = 0$ to time $t = 2$ , we need to integrate $|v(t)|$ . To remove the absolute value bars, we analyze the sign of the velocity function, $v(t) = 48-32t$ . Setting $v(t) = 0$ , we find that $t = 3/2$ seconds. For $t < 3/2$ , we have $v(t) > 0$ and for $t > 3/2$ , we have $v(t) < 0$ .
Thus, $|v(t)| = \left \{ \begin {array}{rc} v(t) & \ \text {if} \;\, t < 3/2 \\ -v(t) & \ \text {if} \;\, t > 3/2 \end {array} \right .$ The distance traveled from $t= 0$ to $t = 2$ , is given by \begin{align*} \int _0^2 |v(t)| \ dt &= \int _0^{3/2} |v(t)| \ dt + \int _{3/2}^2 |v(t)| \ dt\\ &= \int _0^{3/2} v(t) \ dt - \int _{3/2}^2 v(t) \ dt \end{align*}

We will compute these integrals separately. The first integral is \begin{align*} \int _0^{3/2} v(t) \ dt & = \int _0^{3/2}(48 - 32t) \ dt \\ & = (48t - 16t^2)\bigg |_0^{3/2} \\ & = 36 \text{ feet} \end{align*}

The second integral is \begin{align*} \int _{3/2}^2 v(t) \ dt & = \int _{3/2}^2(48 - 32t) \ dt \\ & = (48t - 16t^2)\bigg |_{3/2}^2 \\ & = 32 - 36 = -4 \text{ feet} \end{align*}

Subtracting gives the total distance traveled: \begin{align*} \int _0^2 |v(t)| \ dt & = \int _0^{3/2} v(t) \ dt - \int _{3/2}^2 v(t) \ dt \\ & = 36 - (-4) = 40 \text{ feet}. \end{align*}

(problem 1a) An object is launched vertically with a velocity of $64 ft/sec$ . Assume $g = -32 \ ft/sec^2$ .

(a): Find the displacement of the object from time $t = 0$ to time $t = 3$ .
(b): Find the total distance traveled from time $t = 0$ to time $t = 3$ .

$\text {Displacement} =\answer {48} \ ft,$
$\text {Distance traveled} = \answer {80} \ ft.$

Here is a video solution to problem 1a.

(problem 1b) An object is launched vertically from a height of $5 ft$ with a velocity of $16 ft/sec$ . Assume $g = -32 \ ft/sec^2$ .

(a): Find a formula, $s(t)$ , for the height of the object at time $t$ seconds.
(b): Find the maximum height reached by the object.

$s(t) = \answer {-16t^2 + 16t + 5} feet.$
$\text {Maximum height} =\answer {9} \ feet.$

Here is a video solution to problem 1b.

2026-04-23 13:42:09