Question

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

Answer 1

Answer

The acceleration is

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

Explanation

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

$a=-2.46667ms^{-2}$