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3 years ago

A car is traveling at a velocity of 20 m/s. the driver applies the brakes, and the car slows with an acceleration of a = -4.0 me

ters per second squared. what is the stopping difference of the car?

Physics

1 answer:

sasho [114]3 years ago

3 0

$\displaystyle\\
V = a\times t\\\\
t= \frac{V}{a} = \frac{20}{4} = 5~s\\\\
D = \frac{a\times t^2}{2} = \frac{4\times 5^2}{2} = 2 \times 25 = 50~m$

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Complete Question

The complete question is shown on the uploaded image

Answer:

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Explanation:

From the question we are told that

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