d given by a = 2.82 - 0.003v², velocity in feet per second
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![Question :)
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 = 2.82 - 0.003v2,
where v is the velocity in feet per second.
Determine the velocity vg at the bottom B of the
grade.
755'
B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F413da018-27a8-4aba-8ac6-ae443113d8d5%2Ff46198fb-46c6-4073-8562-889c5512848d%2Fp0y45wk_processed.jpeg&w=3840&q=75)
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