Chapter 5, Problem 48P

(a)

To determine

To Calculate: The minimum distance to make the car stop.

Answer to Problem 48P

The minimum distance it takes to stop the car is $49.1\text{m}$

Explanation of Solution

Given Information:

An automobile is going up a ${15}^{\text{o}}$ grade at a speed of $30\text{ m/s}$ .

The coefficient of static friction between the tires and the road is $0.70$ .

Formula Used:

Third equation of motion:

  $v_1^2=v_0^2+2as$

Where symbols have their usual meanings.

Calculation:

Initial velocity of the automobile $v_0=30\text{m/s}$

Final velocity of the automobile $v_1=0\text{m/s}$

Let the minimum displacement be $s_{\min }$

Let the acceleration of the car be $a$

Free body diagram for automobile going up the inclined plane is shown below

  

Use the third kinematics relation

  $\begin{array}{l}{v}_1^2=v_0^2+2a{s}_{\min }\\ {\left(0\text{ m/s}\right)}^2=v_0^2+2a{s}_{\min }\\ s_{\min }=\frac{-v_0^2}{2a}\mathrm{..............}\left(1\right)\end{array}$

From the free body diagram, along the x-axis

  $\begin{array}{l}{f}_s+mg\sin {15}^{\text{o}}=-ma\\ {\mu }_s{F}_n+mg\sin {15}^{\text{o}}=-ma\mathrm{...}\left(2\right)\end{array}$

Along the y-axis:

  $\begin{array}{l}{F}_n-mg\cos {15}^{\text{o}}=0\\ F_n=mg\cos {15}^{\text{o}}\mathrm{...}\left(3\right)\end{array}$

Substitute $\left(3\right)\text{ in }\left(2\right)$

  $\begin{array}{l}{\mu }_s\left(mg\cos {15}^{\text{o}}\right)+mg\sin {15}^{\text{o}}=-ma\\ a=-g\left({\mu }_s\cos {15}^{\text{o}}+\sin {15}^{\text{o}}\right)\\ =-\left(9.81{\text{ m/s}}^2\right)\left(\left(0.7\right)\cos {15}^{\text{o}}+\sin {15}^{\text{o}}\right)\\ =-9.17{\text{ m/s}}^2\end{array}$

Substitute the value of acceleration in $\left(1\right)$

  $\begin{array}{c}{s}_{\min }=\frac{-v_0^2}{2a}\\ =\frac{-{\left(30\text{ m/s}\right)}^2}{2\left(-9.17{\text{ m/s}}^2\right)}\\ =49.1\text{ m}\end{array}$

Conclusion:

Thus, the minimum stopping distance is $49.1\text{ m}\text{.}$

(b)

To determine

To Calculate: The minimum distance it would take to stop if the car were going down the grade.

Answer to Problem 48P

  $110\text{ m}$

Explanation of Solution

Given Information:

An automobile is going up a ${15}^{\text{o}}$ grade at a speed of $30\text{ m/s}$ .

The coefficient of static friction between the tires and the road is $0.70$ .

FormulaUsed:

Third equation of motion:

  $v_1^2=v_0^2+2as$

Where, symbols have their usual meanings

Calculation:

Initial velocity of the car $v_0=30\text{m/s}$

Final velocity of the car $v_1=0\text{m/s}$

Let the minimum displacement be $s_{\min }$

And let the acceleration of the car be $a$

Use the third kinematic equation

  $\begin{array}{l}{v}_1^2=v_0^2+2a{s}_{\min }\\ {\left(0\text{ m/s}\right)}^2=v_0^2+2a{s}_{\min }\\ s_{\min }=\frac{-v_0^2}{2a}\mathrm{...}\left(4\right)\end{array}$

Along the x-axis

  $\begin{array}{l}{f}_s-mg\sin {15}^{\text{o}}=ma\\ {\mu }_s{F}_n-mg\sin {15}^{\text{o}}=ma\mathrm{...}\left(5\right)\end{array}$

And along the y-axis we conclude that,

  $\begin{array}{l}{F}_n-mg\cos {15}^{\text{o}}=0\\ F_n=mg\cos {15}^{\text{o}}\mathrm{....}\left(6\right)\end{array}$

Substitute $\left(6\right)\text{ in }\left(5\right)$

  $\begin{array}{l}{\mu }_s\left(mg\cos {15}^{\text{o}}\right)-mg\sin {15}^{\text{o}}=ma\\ a=g\left({\mu }_s\cos {15}^{\text{o}}-\sin {15}^{\text{o}}\right)\\ =\left(9.81{\text{ m/s}}^2\right)\left(\left(0.7\right)\cos {15}^{\text{o}}-\sin {15}^{\text{o}}\right)\\ =4.09{\text{ m/s}}^2\end{array}$

Substitute the value of acceleration in $\left(1\right)$

  $\begin{array}{c}{s}_{\min }=\frac{-v_0^2}{2a}\\ =\frac{-{\left(30\text{ m/s}\right)}^2}{2\left(\text{4}{\text{.09 m/s}}^2\right)}\\ =-110\text{ m}\end{array}$

Here, the negative sign indicates that the displacement of the car is in the opposite direction as in part $\left(A\right)$ .

Conclusion:

Therefore, the magnitude of displacement is $110\text{ m}$ .