The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. a) if the car’s speed decreases at a constant rate from 74 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? answer is in mi/h^2b) what distance does the car travel during the braking period? answer is in ft

Acceleration is -30000 mi/h²

Distance travelled in the 3 seconds of deceleration is 261.888 feet

Explanation:

t = Time taken for the car to slow down = 3 s =

u = Initial velocity = 75 mi/h

v = Final velocity = 50 mi/h

s = Displacement

a = Acceleration

Equation of motion

Acceleration is -30000 mi/h²

Converting to feet

1 mile = 5280 feet

0.0496 mile = 0.0496×5280 = 261.888 feet

Distance travelled in the 3 seconds of deceleration is 261.888 feet

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