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A car coasts (engine off) up a 30 grade. If the speed of the car is 27 m/s at the bottom of the grade, what is the distance t
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Answer #1

Initial velocity u = 27 m/s

Final velocity v = 0 m/s

No acceleration from the car so net acceleration

a = -gsin30 = -g/2

V2 = u2 + 2ah

0 = u22 + 2(-g/2)h

gh = u2

h = u2/g

h = 272 / 9.81 = 74.3 m

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