# If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the...

## Question:

If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 25.1 m/s?

## Acceleration:

Acceleration is a physical quantity that defines the difference between the initial and the final speed of a body in motion within a time interval. Acceleration is typically measured in units of meters per second squared.

## Answer and Explanation:

**Given data**:

- The coefficient of kinetic friction is {eq}\mu = 0.84 {/eq}

- The final speed is {eq}v = 25.1\,{\rm{m/s}} {/eq}

The coefficient of friction describes the ratio of the maximum force due to friction on the surface to the normal force applied by the surface, which is equal to the weight of the car, so that

{eq}F_{friction} = ma = - \mu mg\\ \Rightarrow a = - \mu \times g {/eq}

Here the acceleration is taken to be in the negative direction as it serves to slow the car down, and thus is in the direction opposite to the vehicle's motion.

Substituting the values in the above equation as,

{eq}\begin{align*} a &= - \mu \times g\\ &= - 0.84 \times 9.8\\ &= - 8.232\,{\rm{m/}}{{\rm{s}}^2} \end{align*} {/eq}

We then find the shortest distance with the maximum deceleration due to friction as

{eq}\begin{align*} {v^2} - {u^2} &= 2ad\\ d &= \dfrac{{{v^2} - {u^2}}}{{2a}} \end{align*} {/eq}

- Here {eq}v = 0\,{\rm{m/s}} {/eq} is the final velocity.

Substituting the values in the above equation gives

{eq}\begin{align*} d &= \dfrac{{{v^2} - {u^2}}}{{2a}}\\ &= \dfrac{{{{\left( 0 \right)}^2} - {{\left( {25.1} \right)}^2}}}{{2 \times \left( { - 8.232} \right)}}\\ &\approx 38\,{\rm{m}} \end{align*} {/eq}

Thus the shortest distance is approximately 38 meters.

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