Question

For an object whose velocity in ft/sec is given by v(t) = cos(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5?

Answer 1

The total distance traveled is given by

$\displaystyle\int_1^5|v(t)|\,\mathrm dt$

You have

$\displaystyle\int_1^5|\cos t|\,\mathrm dt=\int_1^{\pi/2}\cos t\,\mathrm dt-\int_{\pi/2}^{3\pi/2}\cos t\,\mathrm dt+\int_{3\pi/2}^5\cos t\,\mathrm dt$
$=\left(\sin\dfrac\pi2-\sin1\right)-\left(\sin\dfrac{3\pi}2-\sin\dfrac\pi2\right)+\left(\sin5-\sin\dfrac{3\pi}2\right)$
$=(1-\sin1)-(-1-1)+(\sin5-(-1))$
$=4-\sin1+\sin5$
$\approx2.1996\text{ ft}$