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Mrs. Botts applies brakes on a car help it to decelerate at the rate of -0.80m/s2. What distance is required to stop the car when it is moving 17 m/s.

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ninarAraz

Answer:

Explanation:

vf=0

vi = 17

a = -0.8

Δx = ?

vf^2 = vi^2 + 2 a Δx

Δx = (vf^2 - vi^2) / 2a

Δx = (0-17^2) / 2 (-0.8)

Δx = 180.625 m

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