velocity vs distance

Question

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

0.45

0.26

2.82

0.03

Answer 1

Gail,

 

To do this, you must integrate v(t) from t = 1 to t = 5. This is true because the definite integral of velocity is the net change in distance (which is called displacement).

 

So,

 

∫ sin t dt from t = 1 to t = 5  =  [-cos t]_1^5 = (-cos 5 - (-cos(1)) = cos 1 - cos 5.

 

There is, unfortunately, no way to write this answer exactly. If you need an approximation, you need to use a calculator (in RADIANS) mode.