Chapter 2, Problem 73AP

(a)

To determine

Time at which Kathy overtakes Stan.

Answer to Problem 73AP

Kathy overtakes Stan at $\underset{\_}{5.46\text{s}}$.

Explanation of Solution

Kathy leaves the starting line $1.00\text{s}$ after the start of Stan. Let $t_k$ be the time of travel of Kathy and $t_s$ be the time of travel of Stan.

Write the expression for $t_s$.

    $t_s=t_K+1.00\text{s}$.                                                                                                           (I)

Here, $t_K$ is the time of travel of Kathy’s car, and $t_s$ is the time of travel of Stan’s car.

Write the expression for the final position of Kathy’s car.

    $x_K=v_{xi,K}{t}_K+\frac{1}{2}{a}_{x,K}{t}_K^2$                                                                                              (II)

Here, $x_K$ is the final position of Kathy’s car, $v_{xi,K}$ is the initial velocity of Kathy’s car, and $a_{x,K}$ is the acceleration of Kathy’s car.

Write the expression for the final position of Stan’s car.

    $x_s=v_{xi,s}{t}_s+\frac{1}{2}{a}_{x,s}{t}_s^2$                                                                                               (III)

Here, $x_s$ is the final position of Stan’s car, $v_{xi,s}$ is the initial velocity of Stan’s car, and $a_{x,s}$ is the acceleration of Stan’s car.

Since both the cars start from rest, their initial velocities are zero. That is $v_{xi,K}\text{and }{v}_{xi,s}$ are zero.

    $x_K=\frac{1}{2}{a}_{x,K}{t}_K^2$                                                                                                          (IV)

    $x_s=\frac{1}{2}{a}_{x,s}{t}_s^2$                                                                                                             (V)

Use expression (I) in (V) to get $x_s$.

    $x_s=\frac{1}{2}{a}_{x,s}{\left(t_K+1.00\text{s}\right)}^2$                                                                                           (VI)

When Kathy’s car overtakes Stan’s car, the position of Stan’s car is equal to position of Kathy’s car. Therefore equate expressions (V) and (VI).

    $\begin{array}{c}\frac{1}{2}{a}_{x,s}{\left(t_K+1.00\text{s}\right)}^2=\frac{1}{2}{a}_{x,K}{t}_K^2\\ a_{x,s}{t}_K^2+2a_{x,s}{t}_K+a_{x,s}=a_{x,K}{t}_K^2\\ t_K^2\left(a_{x,K}-a_{x,s}\right)-2a_{x,s}{t}_K-a_{x,s}=0\end{array}$                                                                 (VII)

Expression (VII) is a quadratic equation.

Write the expression for finding the solution of (VII).

    $t_K=\frac{-2a_{x,s}\pm \sqrt{{\left(2a_{x,s}\right)}^2-4\left(a_{x,K}-a_{x,s}\right)\left(-a_{x,s}\right)}}{2\left(a_{x,K}-a_{x,s}\right)}$                                                    (VIII)

Conclusion:

Substitute $3.50{\text{m/s}}^2$ for $a_{x,s}$, and $4.90{\text{m/s}}^2$ for $a_{x,K}$ in equation (VIII) to find $t_K$.

    $\begin{array}{c}{t}_K=\frac{2\left(3.50{\text{m/s}}^2\right)\pm \sqrt{{\left(2\left(3.50{\text{m/s}}^2\right)\right)}^2-4\left(4.90{\text{m/s}}^2-3.50{\text{m/s}}^2\right)\left(-3.50{\text{m/s}}^2\right)}}{2\left(4.90{\text{m/s}}^2-3.50{\text{m/s}}^2\right)}\\ =5.46\text{s}\end{array}$

Therefore, Kathy overtakes Stan at $\underset{\_}{5.46\text{s}}$.

(b)

To determine

The distance travelled by Kathy before she catches Stan.

Answer to Problem 73AP

The distance travelled by Kathy before she catches Stan is $\underset{\_}{73.0\text{m}}$.

Explanation of Solution

Use expression (IV) to find the distance travelled by Kathy before catching him.

    $x_K=\frac{1}{2}{a}_{x,K}{t}_K^2$

Conclusion:

Substitute $4.90{\text{m/s}}^2$ for $a_{x,K}$ and $5.46\text{s}$ for $t_K$ in above equation to find $x_K$.

    $\begin{array}{c}{x}_K=\frac{1}{2}\left(4.90{\text{m/s}}^2\right){\left(5.46\text{s}\right)}^2\\ =73.0\text{m}\end{array}$

Therefore, the distance travelled by Kathy before she catches Stan is $\underset{\_}{73.0\text{m}}$.

(c)

To determine

The speed of both cars at the instant of overtaking of Stan by Kathy.

Answer to Problem 73AP

The speed of Kathy’s car is $\underset{\_}{26.7\text{m/s}}$ and speed of Stan is $\underset{\_}{22.6\text{m/s}}$.

Explanation of Solution

Write the expression for the final velocity of Kathy.

    $v_{xf,K}=v_{xi,K}+a_{x,K}{t}_K$                                                                                                 (IX)

Here, $v_{xf,K}$ is the final velocity of Kathy’s car, $v_{xi,K}$ is the initial velocity of Kathy’s car.

Write the expression for the final velocity of Stan.

    $v_{xf,s}=v_{xi,s}+a_{x,s}{t}_s$                                                                                                     (X)

Here, $v_{xf,s}$ is the final velocity Stan’s car, $v_{xi,s}$ is the initial velocity of Stan’s car.

Conclusion:

Substitute $4.90{\text{m/s}}^2$ for $a_{x,K}$, $0\text{m/s}$ for $v_{xi,K}$, and $5.46\text{s}$ for $t_K$ in equation (IX) to find $v_{xf,K}$.

    $\begin{array}{c}{v}_{xf,K}=\left(4.90{\text{m/s}}^2\right)\left(5.46\text{s}\right)\\ =26.7\text{m/s}\end{array}$

Substitute $5.46\text{s}$ for $t_K$ in equation (I) to find $t_s$.

    $\begin{array}{c}{t}_s=5.46\text{s}+1.00\text{s}\\ =\text{6}\text{.46}\text{s}\end{array}$

Substitute $3.50{\text{m/s}}^2$ for $a_{x,s}$, $0\text{m/s}$ for $v_{xi,s}$, and $6.46\text{s}$ for $t_s$ in equation (X) to find $v_{xf,s}$.

    $\begin{array}{c}{v}_{xf,s}=\left(3.50{\text{m/s}}^2\right)\left(6.46\text{s}\right)\\ =22.6\text{m/s}\end{array}$

Therefore, the speed of Kathy’s car is $\underset{\_}{26.7\text{m/s}}$ and speed of Stan is $\underset{\_}{22.6\text{m/s}}$.