Chapter 3, Problem 46P

(a)

To determine

Distance required to bring the car to a stop when the car is skidding.

Answer to Problem 46P

The distance is $\underset{\_}{54\text{m}}$.

Explanation of Solution

According to Newton’s laws

    $a=-{\mu }_{\text{K}}g$        (I)

Here ${\mu }_{\text{K}}$ is the coefficient of kinetic friction and $g$ is the acceleration due to gravity.

The acceleration is in the opposite direction from the velocity, and so has a negative sign.

Then the distance travelled is,

    $\begin{array}{c}\Delta x_{skid}=\frac{v_0^2}{2a}\\ =-\frac{v_0^2}{2\left(-{\mu }_{\text{K}}g\right)}\\ =\frac{v_0^2}{2{\mu }_{\text{K}}g}\end{array}$        (II)

Here $v_0$ is the velocity.

Conclusion:

Substitute $\text{65}\text{mi/h}$ for $v_0$, $0.80$ for ${\mu }_{\text{K}}$ and $9.8\text{m}\text{}\text{/}\text{}{\text{s}}^2$ for $g$ in (II)

    $\begin{array}{c}\Delta x_{\text{skid}}=\frac{{\left(\text{65}\text{mi/h}\left(\frac{0.44704\text{m/s}}{1\text{mi/h}}\right)\right)}^2}{2(0.80)(9.8\text{m}\text{}\text{/}\text{}{\text{s}}^2)}\\ =\text{54 m }\\ \approx \text{ 180 ft}\end{array}$

The distance is $\underset{\_}{54\text{m}}$.

(b)

To determine

Distance required to bring the car to a stop when the wheels are not locked up.

Answer to Problem 46P

The distance is $\underset{\_}{48\text{m}}$.

Explanation of Solution

In the case of non-skidding, the distance covered would be

    $\Delta x_{\text{no}\text{skid}}=\frac{v_0^2}{2{\mu }_{s}g}$

Here ${\mu }_s$ is the coefficient of kinetic friction

Conclusion:

Substitute $\text{65}\text{mi/h}$ for $v_0$, $0.90$ for ${\mu }_s$ and $9.8\text{m}\text{}\text{/}\text{}{\text{s}}^2$ for $g$ in (II)

    $\begin{array}{c}\Delta x_{\text{no}}{}_{\text{skid}}=\frac{{\left(\text{65}\text{mi/h}\left(\frac{0.44704\text{m/s}}{1\text{mi/h}}\right)\right)}^2}{2(0.90)(9.8\text{m}\text{}\text{/}\text{}{\text{s}}^2)}\\ =48\text{ m }\\ \approx \text{ 160 ft}\end{array}$

The distance is $\underset{\_}{48\text{m}}$.

(c)

To determine

The distance the wheels go if the wheels lock into a skidding stop.

Answer to Problem 46P

The distance is $\underset{\_}{\text{5}\text{.9}\text{m}}$.

Explanation of Solution

The skidding car goes by

    $\begin{array}{c}\Delta x_{\text{skid}}-\Delta x_{\text{no}\text{skid}}=53.6\text{m}-4\text{7}\text{.7 m}\\ =5.\text{9}\text{m}\end{array}$

Or,

    $\begin{array}{c}\frac{\Delta x_{\text{skid}}}{\Delta x_{\text{no}\text{skid}}}=\frac{53.6\text{m}}{\text{47}\text{.7 m}}\\ =1.13\end{array}$

The ratio implies that at any speed, the skidding car will have $13\%$ higher stopping distance. The antilock brakes allow the wheels to keep turning the car is accelerated by kinetic friction.

The stopping distance with antilock brakes is shorter.

Conclusion:

The distance is $\underset{\_}{\text{5}\text{.9}\text{m}}$.

(d)

To determine

Whether antilock brake makes big difference in emergency stops.

Answer to Problem 46P

The stopping distance of cars with antilock brakes are significantly shorter around $13\%$ and the car is safer during emergency stop. Yes, it makes a difference.

Explanation of Solution

Antilock brakes generally offers improved vehicle control and decreases stopping distances on dry and some slippery surfaces, on loose gravel or snow-covered surfaces. They significantly decrease braking distance, while still improving steering control.

Conclusion:

Therefore, the stopping distance of cars with antilock brakes are significantly shorter around $13\%$ and the car is safer during emergency stop. Yes, it makes a difference.