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A car traveling at 26.8m/s hits the brakes to slow down before turning into a neighborhood. Determine how fast it is going after slowing down at a rate of -8.2m/s2 for 14.3seconds.

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Answer:

The speed of the car after the break was applied is  90.46 m/s

Explanation:

Given;

initial velocity of the car, u = 26.8 m/s

accelertation of the car, a = -8.2 m/s²

time of motion, t = 14.3 s

final velocity, v = ?

Apply the following kinematic equation;

v = u + at

v = 26.8 + (-8.2 x 14.3)

v = 26.8 -  117.26

v = -90.46 m/s

magnitude of the final velocity = 90.46 m/s

Therefore, the speed of the car after the break was applied is  90.46 m/s

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