The driver of a car, which is initially at rest at the top a of the grade, releases the brakes and coasts down the grade with an acceleration in feet per second squared given by a = 3.22-0.004v2 , where v is the velocity in feet per second. determine the velocity vb at the bottom b of the grade.

28.3 ft/s

Explanation:

We are given that

Initial velocity of a car=

Acceleration of the car=

We have to find the velocity at the bottom of the grade.

Distance covered by car=600 ft

We know that

Acceleration=a=

Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to

Using substitution method  and

Hence, the velocity at the bottom of the grade=28.3 ft/s

where are the vectors ?