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\frac\pi2-\sin1\right)-\left(\sin\frac{3\pi}2-\sin\frac\pi2\right)+\left(\sin5-\sin\frac{3\pi}2\right)`

`=(1-\sin1)-(-1-1)+(\sin5-(-1))`

`=4-\sin1+\sin5`

`\approx2.1996\text{ ft}`