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

The acceleration is

`a=-2.5ms^(-2)` to the nearest tenth

Step-by-step 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=-(37)/(15)ms^(-2)`

or

`a=-2.46667ms^(-2)`