Chapter 8, Problem 37P

To determine

Find the acceleration required to reach the speed of 10 mph from the rest position within the time period of 25 seconds. Calculate the total distance traveled by bicyclist and also calculate the average speed of cyclist during the first 25 s, 620 s, and 625 s.

Explanation of Solution

Calculate the acceleration required to reach the speed of 10 mph from the rest position within 25 s.

$\begin{array}{c}\text{acceleration}=\frac{\text{change}\text{in}\text{velocity}}{\text{time}}\\ =\frac{10\frac{\text{miles}}{\text{h}}}{25\text{s}}\\ =\frac{10\frac{\text{miles}}{\text{h}}\left(\frac{1\text{h}}{3600\text{s}}\right)}{25\text{s}}\left\{\because 1\text{h}=3600\text{s}\right\}\\ =1.11\times {10}^{-4}\frac{\text{m}}{{\text{s}}^2}\end{array}$

During the initial 25 seconds, the speed increases from 0 to 10 mph. Therefore, the average speed is $\frac{10\text{mph}}{2}=5\text{mph}$.

Calculate the distance traveled by the bicyclist with the average speed of 5 miles/h during the initial 25 s as follows.

$\begin{array}{c}{d}_1=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(25\text{s}\right)\left(5\frac{\text{miles}}{\text{h}}\right)\\ =\left(25\text{s}\right)\left(5\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =183.33\text{ft}\end{array}$

Consider the constant average speed of 10 miles/h in the next 10 minutes.

Calculate the distance traveled by the bicyclist with the average speed of 10 miles/h during the next 10 minutes.

$\begin{array}{c}{d}_2=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(600\text{s}\right)\left(10\frac{\text{miles}}{\text{h}}\right)\\ =\left(600\text{s}\right)\left(10\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =8800\text{ft}\end{array}$

Here, the bicyclist decelerates at the constant rate from a speed of 10 mph to 0, the average speed of the bicyclist during the last 5 seconds is 5 mph.

Calculate the distance traveled during this period as below.

$\begin{array}{c}{d}_3=\left(\text{time}\right)\left(\text{average}\text{speed}\right)\\ =\left(5\text{s}\right)\left(5\frac{\text{miles}}{\text{h}}\right)\\ =\left(5\text{s}\right)\left(5\frac{\text{miles}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left(\frac{5280\text{ft}}{1\text{mile}}\right)\left\{\begin{array}{l}\because 1\text{h}=3600\text{s,}\text{and}\\ 1\text{mile}=5280\text{ft}\end{array}\right\}\\ =36.66\text{ft}\end{array}$

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

Substitute 1833.33 ft for $d_1$, 8800 ft for $d_2$, and 36.66 ft for $d_3$,

$\begin{array}{c}d=183.33\text{ft}+\text{8800}\text{ft}+36.66\text{ft}\\ =9020\text{ft}\\ =\text{9020}\text{ft}\left(\frac{1\text{mile}}{\text{5280}\text{ft}}\right)\left\{\because 1\text{mile}=\text{5280}\text{ft}\right\}\\ =\text{1}\text{.71}\text{miles}\end{array}$

Use the expression for the average speed of the bicyclist for the entire duration of travel.

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

Substitute 9020 feet for distance traveled and 625 s for time.

$\begin{array}{c}{V}_{\text{average}}=\frac{9020\text{ft}}{625\text{s}}\\ =14.43\frac{\text{ft}}{\text{s}}\\ =14.43\frac{\text{ft}}{\text{s}}\left(\frac{1\text{mile}}{\text{5280}\text{ft}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\left\{\begin{array}{l}\because 1\text{mile}=\text{5280}\text{ft, and}\\ 1\text{h}=3600\text{s}\end{array}\right\}\\ =9.84\text{mph}\end{array}$

Conclusion:

Thus, the acceleration required to reach the speed of 10 mph in 25 s is $\underset{\_}{1.11\times {10}^{-4}\frac{\text{m}}{{\text{s}}^2}}$, the total distance traveled by bicyclist is $\underset{\_}{\text{1}\text{.71}\text{miles}}$ and the average speed of cyclist during first 25 s, 620 s, and 625 s are $\underset{\_}{5\frac{\text{miles}}{\text{h}},10\frac{\text{miles}}{\text{h}},\text{and}5\frac{\text{miles}}{\text{h}}}$ respectively.