Chapter 2, Problem 38P

(a)

To determine

The time interval during which the bicycle is ahead of the car.

Answer to Problem 38P

The time interval during which the bicycle is ahead of the car is $3.45\text{ s}$.

Explanation of Solution

Write the kinematics equation.

  $v_f=v_i+at$

Here, $v_f$ is the final speed, $v_i$ is the initial speed, $a$ is the acceleration and $t$ is the time interval.

Rewrite the above equation for $t$.

  $t=\frac{v_f-v_i}{a}$

Use the above equation to write the expression for the time taken by the bicycle to reach its maximum speed.

  $t_{b,1}=\frac{v_{b,\text{max}}-v_{b,i}}{a_b}$        (I)

Here, $t_{b,1}$ is the time taken by the bicycle to reach its maximum speed, $v_{b,\text{max}}$ is the maximum speed of the bicycle, $v_{b,i}$ is the initial speed of the bicycle and $a_b$ is the acceleration of the bicycle.

It is given that the acceleration of the car is less than the acceleration of the bicycle. This implies the car cannot catch the bicycle until the time bicycle is at its maximum speed and coasting and it will be greater than the time taken by the bicycle to reach its maximum speed.

Write the equation for the total displacement of the bicycle.

  $\Delta x_b=\frac{1}{2}{a}_b{t}_{b,1}^2+v_{b,\text{max}}\left(t-t_{b,1}\right)$        (II)

Here, $\Delta x_b$ is the total displacement of the bicycle and $t$ is the time taken by the car to reach the bicycle.

It is given that the car starts from rest so that its initial velocity will be zero.

Write the equation for the displacement of the car during the time taken by the bicycle to reach the car.

  $\Delta x_c=\frac{1}{2}{a}_c{t}^2$        (III)

Here, $\Delta x_c$ is the total displacement of the car during time $t$ and $a_c$ is the acceleration of the car.

When the car catches the bicycle, the two displacements will be equal.

Write the condition when the car to catches the bicycle.

  $\Delta x_c=\Delta x_b$        (IV)

Conclusion:

Substitute $20.0\text{ mi/h}$ for $v_{b,\text{max}}$, $0$ for $v_{b,i}$ and $13.0\text{ mi/h}\cdot \text{s}$ for $a_b$ in equation (I) to find $t_{b,1}$.

  $\begin{array}{c}{t}_{b,1}=\frac{20.0\text{ mi/h}-0}{13.0\text{ mi/h}\cdot \text{s}}\\ =1.54\text{ s}\end{array}$

Substitute $13.0\text{ mi/h}\cdot \text{s}$ for $a_b$, $1.54\text{ s}$ for $t_{b,1}$ and $20.0\text{ mi/h}$ for $v_{b,\text{max}}$ in equation (II) to find the expression for $\Delta x_b$.

  $\begin{array}{c}\Delta x_b=\frac{1}{2}\left(13.0\frac{\text{mi/h}}{\text{s}}\frac{1.47\text{ ft/s}}{1\text{ mi/h}}\right){\left(1.54\text{ s}\right)}^2+\left(20.0\text{ mi/h}\frac{1.47\text{ ft/s}}{1\text{ mi/h}}\right)\left(t-1.54\text{ s}\right)\\ =\left(29.4\text{ ft/s}\right)t-22.6\text{ ft}\end{array}$        (V)

Substitute $9.00\text{ mi/h}\cdot \text{s}$ for $a_c$ in equation (III) to find the expression for $\Delta x_c$.

  $\begin{array}{c}\Delta x_c=\frac{1}{2}\left(9.00\frac{\text{mi/h}}{\text{s}}\frac{1.47\text{ ft/s}}{1\text{ mi/h}}\right)t^2\\ =\left(6.62{\text{ ft/s}}^2\right)t^2\end{array}$        (VI)

Put equations (V) and (VI) in equation (IV).

  $\begin{array}{c}\left(6.62{\text{ ft/s}}^2\right)t^2=\left(29.4\text{ ft/s}\right)t-22.6\text{ ft}\\ t^2-\frac{\left(29.4\text{ ft/s}\right)}{6.62{\text{ ft/s}}^2}t+\frac{22.6\text{ ft}}{6.62{\text{ ft/s}}^2}=0\\ t^2-\left(4.44\text{ s}\right)t+3.42{\text{ s}}^2=0\end{array}$        (VII)

Write the quadratic formula to solve the equation $a{t}^2+bt+c=0$.

  $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$        (VIII)

Equation (VII) is a quadratic equation in $t$ . Find the roots of equation (VII) using the quadratic formula given by (VIII).

  $\begin{array}{c}t=\frac{-\left(-4.44\text{ s}\right)\pm \sqrt{{\left(-4.44\text{ s}\right)}^2-4\left(3.42{\text{ s}}^2\right)}}{2}\\ =3.45\text{ s or 0}.99\text{ s}\end{array}$

Since $t>t_{b,1}$ and the value of $t_{b,1}$ is $1.54\text{ s}$, only physically meaningful root of the above equation is $3.45\text{ s}$.

Therefore, the time interval during which the bicycle is ahead of the car is $3.45\text{ s}$.

(b)

To determine

The maximum distance by which the bicycle leads the car.

Answer to Problem 38P

The maximum distance by which the bicycle leads the car is $10.0\text{ ft}$.

Explanation of Solution

The distance by which bicycle will lead the car will increase as long as the bicycle moves faster than the car or when the speed of the car becomes equal to the maximum speed of the bicycle.

Write the equation for the elapsed time when the bicycle’s lead ceases to increase.

  $t=\frac{v_{b,\text{max}}}{a_c}$        (IX)

Here, $t$ is the elapsed time when the bicycle’s lead ceases to increase.

Write the equation for the lead of the bicycle.

  $\text{lead}=\Delta x_{b,t}-\Delta x_{c,t}$

Here, the subscript $t$ denotes the elapsed time.

Put equations (V) and (VI) in the above equation.

  $\text{lead}=\left[\left(29.4\text{ ft/s}\right)t-22.6\text{ ft}\right]-\left(6.62{\text{ ft/s}}^2\right)t^2$        (X)

Conclusion:

Substitute $20.0\text{ mi/h}$ for $v_{b,\text{max}}$ and $9.00\text{ mi/h}\cdot \text{s}$ for $a_c$ in equation (IX) to find $t$ .

  $\begin{array}{c}t=\frac{20.0\text{ mi/h}}{9.00\text{ mi/h}\cdot \text{s}}\\ =2.22\text{ s}\end{array}$

Substitute $2.22\text{ s}$ for $t$ in equation (X) to find the maximum distance by which the bicycle leads the car.

  $\begin{array}{c}\text{lead}=\left[\left(29.4\text{ ft/s}\right)\left(2.22\text{ s}\right)-22.6\text{ ft}\right]-\left(6.62{\text{ ft/s}}^2\right){\left(2.22\text{ s}\right)}^2\\ =10.0\text{ ft}\end{array}$

Therefore, the maximum distance by which the bicycle leads the car is $10.0\text{ ft}$.