Question

For an object whose velocity in ft/sec is given by v(t) = 2sin(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5?

Answer 1

`-2cos(5)+2cos(1)approx0.5` ft

Explanation:

Distance traveled is the definite integral of velocity.

Furthermore, we're being asked for distance, which is not impacted at all by the direction travelled (IE the velocity being positive or negative), so fortunately, there will be no need to examine if/where `v(t)` changes sign, as we would if we were asked for displacement.

So, the distance traveled from one second to five seconds will be given by

`int_1^5 2sin(t)dt=-2cost|_1^5=-2cos(5)+2cos(1)`