Chapter 8, Problem 28P

To determine

Calculate the total distance traveled by the car and the average speed of the car. Plot the acceleration of car as a function of time.

Answer to Problem 28P

The total distance traveled by car is $\underset{\_}{2.25\text{miles}}$ and the average speed of the car is $\underset{\_}{59.3\text{mph}}$.

Explanation of Solution

Refer to the figure Problem 8.28 in the textbook for the speed versus time characteristics of the car.

During the initial 20 seconds, the speed of the car increases linearly from a value of zero to 60 mph. Therefore, the average speed of the car is 30 mph during this period. The distance traveled during this period is,

$\begin{array}{c}{d}_1=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(20\text{s}\right)\left(30\frac{\text{miles}}{\text{h}}\right)\\ =\left(20\text{s}\right)\left(30\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =880\text{ft}\end{array}$

During the next 20 minutes the car moves with a constant speed of 60 mph and the distance traveled during this period is,

$\begin{array}{c}{d}_2=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(1200\text{s}\right)\left(60\frac{\text{miles}}{\text{h}}\right)\\ =\left(1200\text{s}\right)\left(60\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =105600\text{ft}\end{array}$

Because the car decelerates at a constant rate from a speed of 60 mph to 0 mph, the average speed of the car during the last 10 seconds is also 30 mph, and the distance traveled during this period is,

$\begin{array}{c}{d}_3=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(10\text{s}\right)\left(30\frac{\text{miles}}{\text{h}}\right)\\ =\left(10\text{s}\right)\left(30\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =440\text{ft}\end{array}$

The total distance traveled is $d=d_1+d_2+d_3$.

Substitute 880 feet for $d_1$, $105600\text{ft}$ for $d_2$, and 440 feet for $d_3$,

$\begin{array}{c}d=880\text{ft}+105600\text{ft}+440\text{ft}\\ =\text{106920}\text{ft}\\ =\text{106920}\text{ft}\left(\frac{1\text{mile}}{\text{5280}\text{ft}}\right)\left\{\because 1\text{mile}=\text{5280}\text{ft}\right\}\\ =\text{20}\text{.25}\text{miles}\end{array}$

Give the expression for average speed of the car for the entire duration of travel as below.

$V_{\text{average}}=\frac{\text{distance}\text{traveled}}{\text{time}}$

Substitute $\text{106920}\text{ft}$ for distance traveled and 1230 s for time.

$\begin{array}{c}{V}_{\text{average}}=\frac{106920\text{ft}}{1230\text{s}}\\ =86.9\frac{\text{ft}}{\text{s}}\\ =86.9\frac{\text{ft}}{\text{s}}\left(\frac{1\text{mile}}{\text{5280}\text{ft}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\left\{\begin{array}{l}\because 1\text{mile}=\text{5280}\text{ft, and}\\ 1\text{h}=3600\text{s}\end{array}\right\}\\ =59.3\text{mph}\end{array}$

To plot the acceleration of car as a function of time, calculate the acceleration during the first 20 seconds, the next 20 minutes, and the last 10 seconds.

Calculate the acceleration of the car during the initial 20 seconds as below.

$\begin{array}{c}\text{accelaration}=\frac{88\text{ft/s}}{20\text{s}}\\ =4.4{\text{ft/s}}^2\end{array}$

Note during this 20 seconds period, the speed of car changes from 0 to $V=60\text{mph}$, which is equal to $88\text{ft/s}$.

During the next 20 minutes, the car moves with constant speed of 60 mph. Therefore, the acceleration is zero.

Show the plot of acceleration versus time as in Figure 1.

Conclusion:

Thus, the total distance traveled by car is $\underset{\_}{2.25\text{miles}}$, the average speed of the car is $\underset{\_}{59.3\text{mph}}$, and the plot of acceleration of car as a function of time is drawn.