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ololo11 [35]
3 years ago
15

A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration

Physics
1 answer:
LuckyWell [14K]3 years ago
6 0

<u>Answer</u>

The acceleration is

a=-2.5ms^{-2} to the nearest tenth

<u>Explanation</u>

Since the car was travelling at 37ms^{-1} before it starts to decelerate, the initial velocity is

u=37ms^{-1}.

The final velocity is v=0ms^{-1}, because the car came to a stop.

The time taken is t=15s.

Using the Newton's equation of linear motion,

v=u +at, we find the acceleration by substituting the known values.


This implies that,

0=37 +a(15)

This gives us,

0-37=15a


\Rightarrow -37=15a


We divide both sides by 15 to get,

a=-\frac{37}{15}ms^{-2}

or

a=-2.46667ms^{-2}




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