Question

suppose a certain object moves in a straight line with the following velocity, where v is in meter per second and t is in seconds. v(t) = -2 + t = 3sin(pit). Without using your calculator, but instead, using properties of definite integrals and facts you know about area. determine the net change in distance of the object from  time t=0 to time t=6 seconds.  And find the object's average velocity in this interval.

Answer 1

The given function is
v(t) = -2 + t  - sin (πt)
The next change in the distance from 0 to 6 is
v(0) = 0
v(6) = 4
(4 - 0) / (6 - 0) = 2/3

The average velocity in the interval is
∫v(t) dt from 0 to 6
= -2t + t²/2 + 6πcos πt from 0 to 6
= -2(6) + 36/2 - 6 - (0 + 0 +1)
= -1