Chapter 2, Problem 38P

(a)

To determine

The time taken by the first car to move after the second car begin to pass the first car.

Answer to Problem 38P

The time taken by the first car to move after the second car passes through it is $1.21\text{s}$ .

Explanation of Solution

Given:

The time difference between the two cars is $0.500\text{s}$ .

Formula Used:

The time taken by the first car is $0.500\text{s}$ more than the second car, thus the distance travelled by the first car is given by,

  $s_1=v_{1x0}\left(t+0.65\text{s}\right)+\frac{1}{2}a{\left(t+0.5\text{s}\right)}^2$

The expression for the distance travelled by the second car is given by,

  $s_2=v_{2x0}t+\frac{1}{2}\left(2a\right)t^2$

The expression for the relation between the distance travelled by both the cars is calculated as,

  $s_1=s_2$

Calculation:

The first car starts from the rest, thus the initial velocity of the first car is $v_{1x0}=0\text{m}/\text{s}$ .

The distance travelled by the first car is calculated as,

  $\begin{array}{c}{s}_1=v_{1x0}\left(t+0.65\text{s}\right)+\frac{1}{2}a{\left(t+0.5\text{s}\right)}^2\\ =\left(0\right)\left(t+0.65\text{s}\right)+\frac{1}{2}a{\left(t+0.5\text{s}\right)}^2\\ =\frac{1}{2}a{\left(t+0.5\text{s}\right)}^2\end{array}$

The second car starts from the rest, thus the initial velocity of the second car is $v_{2x0}=0\text{m}/\text{s}$ .

The distance travelled by the second car is calculated as,

  $\begin{array}{c}{s}_2=v_{2x0}t+\frac{1}{2}\left(2a\right)t^2\\ =\left(0\right)t+\frac{1}{2}\left(2a\right)t^2\\ =a{t}^2\end{array}$

The time after which the first car begins to move after the second car passes is given by,

  $\begin{array}{l}{s}_1=s_2\\ \frac{1}{2}a{\left(t+0.5\text{s}\right)}^2=a{t}^2\\ t+0.5=\sqrt{2t}\\ t=1.21\text{s}\end{array}$

Conclusion:

Therefore, the time taken by the first car to move after the second car passes through it is $1.21\text{s}$ .

(b)

To determine

The ratio for the speed of both the car at the time $t=1.21\text{s}$ .

Answer to Problem 38P

The ratio of the final speed of both the car is $1.41$ .

Explanation of Solution

Formula Used:

The expression for the final velocity of the first car is given by,

  $v_{1x}=v_{1x0}+2at$

The expression for the final velocity of the second car is given by,

  $v_2=v_{2x0}+a\left(t+0.5\right)$

Calculation:

The final velocity of the first car is calculated as,

  $\begin{array}{c}{v}_{1x}=v_{1x0}+2at\\ =\left(0\text{m}/\text{s}\right)+2at\\ =2at\end{array}$

The final velocity of the second car is calculated as,

  $\begin{array}{c}{v}_2=v_{2x0}+a\left(t+0.5\right)\\ =0+a\left(t+0.5\right)\\ =a\left(t+0.5\right)\end{array}$

The ratio of the velocity is calculated as,

  $\begin{array}{c}\frac{v_{1x}}{v_{2x}}=\frac{2at}{a\left(t+0.5\right)}\\ =\frac{2a\left(1.21\text{s}\right)}{a\left(\left(1.21\text{s}\right)+0.5\right)}\\ =1.41\end{array}$

Conclusion:

Therefore, the ratio of the final speed of both the car is $1.41$ .