Chapter 3, Problem 36P

(a)

To determine

The instantaneous acceleration of the sports car at time $t=14\text{s}$ from start,

Answer to Problem 36P

The instantaneous acceleration of the sports car at time $t=14\text{s}$ from start is $\underset{\_}{3.92\text{m}/{\text{s}}^2}$.

Explanation of Solution

Figure 1 represents the graph between velocity $v_{\text{x}}$ and time $t$.

The instantaneous acceleration of the car is the slope of the velocity-time graph.

Write the expression for slope at the time $14\text{s}$ from the start using Figure 1.

    $\begin{array}{c}\text{slope =}\frac{y_2-y_1}{x_2-x_1}\\ =\frac{55\text{m}/\text{s}-0\text{m}/\text{s}}{14\text{s}-0\text{s}}\\ =3.92\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the instantaneous acceleration of the sports car at time $t=14\text{s}$ from start is $\underset{\_}{3.92\text{m}/{\text{s}}^2}$.

(b)

To determine

The frictional force of the car at the instant at $t=14\text{s}$ in terms of weight $W$.

Answer to Problem 36P

The frictional force of the car at the instant at $t=14\text{s}$ in terms of weight $W$ is $\underset{\_}{{\mu }_{\text{k}}mg}$.

Explanation of Solution

Write the expression for frictional force.

    $f_{\text{k}}={\mu }_{\text{k}}mg$        (I)

Here, $f_{\text{k}}$ is the kinetic frictional force, ${\mu }_{\text{k}}$ is the coefficient of kinetic frictional force, $m$ is the mass of the car, and $g$ is the acceleration due to gravity.

Conclusion:

Therefore, the frictional force of the car at the instant at $t=14\text{s}$ in terms of weight $W$ is $\underset{\_}{{\mu }_{\text{k}}mg}$.

(c)

To determine

The displacement of the car from $t=12.0\text{s}$ to $t=16.0\text{s}$.

Answer to Problem 36P

The displacement of the car from $t=12.0\text{s}$ to $t=16.0\text{s}$ is $220\text{m}$.

Explanation of Solution

The displacement of the car $\Delta x$ is the area under the graph between velocity and time $t$. Since each box has a height of $5\text{m}/\text{s}$ and width of $2\text{s}$ and there are $22\text{boxes }$ inside the curve in Figure 1, thus the displacement of the car is given by

    $\begin{array}{c}\Delta x=22\left(5\text{m}/\text{s}\right)\left(2\text{s}\right)\\ =220\text{m}\end{array}$

Here, $\Delta x$ is the displacement of the car.

Conclusion:

Therefore, The displacement of the car from $t=12.0\text{s}$ to $t=16.0\text{s}$ is $220\text{m}$.

(d)

To determine

The average velocity of the car in the time interval from $12.0\text{s}$ to $16.0\text{s}$.

Answer to Problem 36P

The average velocity of the car in the time interval from $12.0\text{s}$ to $16.0\text{s}$ is $\underset{\_}{27.5\text{m}/\text{s}}$.

Explanation of Solution

From the graph in Figure 1, for time $t=12.0\text{s}$ the velocity of the car is $52\text{m}/\text{s}$, for time $t=16.0\text{s}$ the velocity of the car is $58\text{m}/\text{s}$.

Write the expression of average velocity.

  $v_{\text{avg}}=\frac{v_{12.0\text{s}}+v_{16.0\text{s}}}{2}$        (II)

Here, $v_{\text{avg}}$ is the average velocity, $v_{12.0\text{s}}$ is the velocity of the car at $12.0\text{s}$, $v_{16.0\text{s}}$ is the velocity of the car at $16.0\text{s}$ and $\Delta t$ is the time interval.

Conclusion:

Substitute $58\text{m}/\text{s}$ for $v_2$, and $52\text{m}/\text{s}$ for $v_1$ in equation (II), to find $v_{\text{avg}}$.

    $\begin{array}{c}{v}_{\text{avg}}=\frac{52\text{m}/\text{s}+58\text{m}/\text{s}}{2}\\ =55\text{m}/\text{s}\end{array}$

Therefore, the average velocity of the car in the time interval from $12.0\text{s}$ to $16.0\text{s}$ is $\underset{\_}{55\text{m}/\text{s}}$.