Question

A car traveling at 13.6 meters per second crashes into a barrier and stops in 0.321 meters.a. How long does it take the car to stop? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

Given,

The initial velocity of the car, u=13.6 m/s

The final velocity of the car, v=0 m/s

The distance covered by the car, d=0.321 m

From the equation of the motion,

`v^2-u^2=2ad`

On rearranging the equation,

`a=(v^2-u^2)/(2d)`

This is the acceleration of the car which brings the car to rest after the collision.

On substituting the known values,

`\begin{gathered} a=(0-13.6^2)/(2*0.321) \\ =-288m/s^2 \end{gathered}`

From another equation of the motion,

`v=u+at`

On rearranging the above equation,

`t=(v-u)/(a)`

On substituting the known values,

`\begin{gathered} t=(0-13.6)/(-288) \\ =0.0472\text{ s} \end{gathered}`

Thus the car comes to stop in 0.0472 seconds.