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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?

User Raphael
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1 Answer

22 votes
22 votes
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}

User Uioporqwerty
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