Chapter 4, Problem 75P

(a)

To determine

The drag racer time and the minimum coefficient of static friction between the tires and the road.

Answer to Problem 75P

The drag racer time is $7.7\text{s}$ and the minimum coefficient of static friction between the tires and the road is $1.38$.

Explanation of Solution

Write the expression for acceleration of the drag racer from Kinematic equation of motion.

    $v_f^2=v_i^2+2as$

Here, $v_f$ is the final velocity of the racer, $v_i$ is the initial velocity of the racer, $a$ is the acceleration of the racer, and $s$ is the distance of the finishing line of the race track.

Rewrite the above expression for acceleration.

$\begin{array}{c}{v}_f^2-v_i^2=2as\\ a=\frac{v_f^2-v_i^2}{2s}\end{array}$        (I)

Write the expression for drag racer time from kinematic equation.

    $v_f=v_i+at$

Here, $t$ is the drag racer time.

Rewrite the above expression for time.

$\begin{array}{c}{v}_f-v_i=at\\ t=\frac{v_f-v_i}{a}\end{array}$        (II)

Write the expression for frictional force.

$f={\mu }_{s}mg$        (III)

Here, $f$ is the frictional force, ${\mu }_s$ is the minimum coefficient of static friction between the tires and road, $m$ is the mass of the drag racer, and $g$ is the acceleration due to gravity.

Write the expression for horizontal force exerted on the racer.

$F=ma$        (IV)

Here, $F$ is the horizontal force and $a$ is the acceleration of the racer.

Compare the above relation (III) and (IV) to find the coefficient of static friction.

$\begin{array}{c}{\mu }_{s}mg=ma\\ {\mu }_{s}g=a\\ {\mu }_s=\frac{a}{g}\end{array}$        (V)

Conclusion:

Substitute $104\text{m/s}$ for $v_f$, $400\text{m}$ for $s$, and $0\text{m/s}$ for $v_i$ in the equation (I) to find $a$.

$\begin{array}{c}a=\frac{{\left(104\text{m/s}\right)}^2-{\left(0\text{m/s}\right)}^2}{2\left(400\text{m}\right)}\\ =13.5{\text{m/s}}^2\end{array}$

Substitute $104\text{m/s}$ for $v_f$, $13.5{\text{m/s}}^2$ for $a$, and $0\text{m/s}$ for $v_i$ in the equation (II) to find $t$.

$\begin{array}{c}t=\frac{104\text{m/s}-0\text{m/s}}{13.5{\text{m/s}}^2}\\ =7.7\text{s}\end{array}$

Substitute $9.80{\text{m/s}}^2$ for $g$ and $13.5{\text{m/s}}^2$ for $a$ in the equation (V) to find ${\mu }_S$.

$\begin{array}{c}{\mu }_s=\frac{13.5{\text{m/s}}^2}{9.80{\text{m/s}}^2}\\ =1.38\end{array}$

Thus, the drag racer time is $7.7\text{s}$ and the minimum coefficient of static friction between the tires and the road is $1.38$.

(b)

To determine

The time taken to finish the race.

Answer to Problem 75P

The time taken to finish the race is $7.76\text{s}$.

Explanation of Solution

If the acceleration is reduced to 30% because of the bad tires or wet pavement, the new acceleration of the drag racer is,

$a^{\prime }=\frac{30}{100}a$        (VI)

Here, $a^{\prime }$ is the new acceleration of the racer.

The net acceleration of the racer is,

$a_{net}=a-a^{\prime }$        (VII)

Write the expression for the time taken to finish the race.

    $t=\frac{v_f}{a_{net}}$        (VIII)

Here, $t$ is the time taken to finish the race

Conclusion:

Substitute $13.5{\text{m/s}}^2$ for $a$ in the equation (VI) to find $a^{\prime }$.

$\begin{array}{c}{a}^{\prime }=\frac{30}{100}\left(13.5{\text{m/s}}^2\right)\\ =0.135{\text{m/s}}^2\end{array}$

Substitute $0.135{\text{m/s}}^2$ for $a^{\prime }$ and $13.5{\text{m/s}}^2$ for $a$ in the equation (VII) to find $a_{net}$.

$\begin{array}{c}{a}_{net}=13.5{\text{m/s}}^2-0.135{\text{m/s}}^2\\ =13.4{\text{m/s}}^2\end{array}$

Substitute $104\text{m/s}$ for $v_f$ and $13.4{\text{m/s}}^2$ for $a_{net}$ in the equation (VIII) to find $t$.

$\begin{array}{c}t=\frac{104\text{m/s}}{13.4{\text{m/s}}^2}\\ =7.76\text{s}\end{array}$

Thus, the time taken to finish the race is $7.76\text{s}$.