Question

An automobiles wheels are locked as it slides to a stop from 28.9m/s. If the coefficient of kinetic friction is 0.224 and the road is horizontal, how long does it take the car to stop? And how far does it travel while stopping?

Answer 1

Answer:

It takes the car 13.2 seconds to stop

The car travels 19.2 m while stopping

Step-by-step explanation:

The coefficient of kinetic friction, μ = 0.224

The initial velocity, u = 28.9 m/s

The final velocity, v = 0 m/s

The force acting in the x-direction is:

`\begin{gathered} \sum ^{}_{}f=ma_x \\ \mu_kmg=ma_x \\ 0.224m(9.8)=ma_x \\ a_x=0.224(9.8) \\ a_x=2.1952m/s^2 \end{gathered}`

The time taken for the car to stop can be calculated using the formula:

`\begin{gathered} v=u+a_xt \\ 0=28.9-2.1952t \\ t=(28.9)/(2.1952) \\ t=13.2\text{ seconds} \end{gathered}`

The distance travelled

`\begin{gathered} S=ut+(1)/(2)at^2 \\ S=28.9(13.2)+0.5(-2.1952)(13.2^2) \\ S=190.2m \end{gathered}`