Chapter 2, Problem 39P

(a)

To determine

The minimum possible distance the person can be away from the light when the light turns yellow.

Answer to Problem 39P

The distance needed by the car to stop safely is $\text{50}\text{m}$ .

Explanation of Solution

Given:

The reaction time of the driver is $t=0.500\text{s}$ .

The initial speed of the car is $v_{1x}=20.0\text{m}/\text{s}$ .

The maximum possible rate of deceleration is $a_x=-5.00\text{m}/{\text{s}}^2$ .

Formula Used:

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

  $v_x^2=v_{x0}^2+2a_{x}x$

The distance travelled by the car for the reaction time $t$ is given by

  $x_t=v_{x0}t$

Calculation:

The car comes to complete stop after reaching the light thus the value of the final velocity is $v_x=0\text{m}/\text{s}$ .

The distance travelled by the car is calculated as,

  $\begin{array}{l}{v}_x^2=v_{x0}^2+2a_{x}x\\ 0={\left(20\text{m}/\text{s}\right)}^2+2\left(-5\text{m}/{\text{s}}^2\right)x\\ x=40\text{m}\end{array}$

The distance travelled by the car for the reaction time $t$ is calculated as,

  $\begin{array}{c}{x}_t=v_{x0}t\\ =\left(20\text{m}/\text{s}\right)\left(0.500\text{s}\right)\\ =10\text{m}\end{array}$

The total distance travelled by the car is calculated as,

  $\begin{array}{c}{x}_t+x=10\text{m}+40\text{m}\\ =\text{50}\text{m}\end{array}$

Conclusion:

Therefore, the distance needed by the car to stop safely is $\text{50}\text{m}$ .

(b)

To determine

The maximum distance that must be between the light and the car so that the car must pass the light before the light turns red.

Answer to Problem 39P

The distance up to which the car must be away from light, to make it through the light is $\text{69}\text{.375}\text{m}$ .

Explanation of Solution

Given:

The maximum rate of acceleration after the brakes are applied is $a_x=3.00\text{m}/{\text{s}}^2$ .

The initial speed of the car is $v_{1x}=20.0\text{m}/\text{s}$ .

Formula Used:

The reaction time after the light turns red is $0.5\text{s}$ and the travel distance is of $10\text{m}$ , thus the remaining time to accelerate past the light is $2.5\text{s}$ .

The expression for the distance traveled by the car is given by,

  $x=v_{x0}t+\frac{1}{2}{a}_x{t}^2$

The expression for the total distance required to travel to make it through the light is given by,

  $x_t=x+10\text{m}$

Calculation:

The distance traveled by the car is calculated as,

  $\begin{array}{c}x=v_{x0}t+\frac{1}{2}{a}_x{t}^2\\ =\left(20\text{m}/\text{s}\right)\left(2.5\text{s}\right)+\frac{1}{2}\left(3\text{m}/{\text{s}}^2\right){\left(2.5\text{s}\right)}^2\\ =59.375\text{m}\end{array}$

The total distance required to travel to make it through the light is calculated as,

  $\begin{array}{c}{x}_t=59.375\text{m}+10\text{m}\\ =\text{69}\text{.375}\text{m}\end{array}$

Conclusion:

Therefore, the distance up to which the car must be away from light, to make it through the light is $\text{69}\text{.375}\text{m}$ .

(c)

To determine

The minimum possible distance the person can be away from the light when the light turns yellow.

Answer to Problem 39P

The distance needed by the car to stop safely is $14\text{0}\text{m}$ and maximum distance between the car and the light so that the car can make it through the light before it turns red is $96.875\text{m}$ .

Explanation of Solution

Given:

The reaction time of the driver is $t=0.500\text{s}$ .

The initial speed of the car is $v_{1x}=35\text{m}/\text{s}$ .

The maximum possible rate of deceleration is $a_d=-5.00\text{m}/{\text{s}}^2$ .

The maximum rate of acceleration after the brakes are applied is $a_x=3.00\text{m}/{\text{s}}^2$ .

Formula Used:

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

  $v_x^2=v_{x0}^2+2a_{d}x$

The distance travelled by the car for the reaction time $t$ is given by

  $x_t=v_{x0}t$

The reaction time after the light turns red is $0.5\text{s}$ and the travel distance is of $10\text{m}$ , thus the remaining time to accelerate past the light is $2.5\text{s}$ .

The expression for the distance traveled by the car is given by,

  $x=v_{x0}t+\frac{1}{2}{a}_x{t}^2$

The expression for the total distance required to travel to make it through the light is given by,

  $x_t=x+10\text{m}$

Calculation:

The car comes to complete stop after reaching the light thus the value of the final velocity is $v_x=0\text{m}/\text{s}$ .

The distance travelled by the car is calculated as,

  $\begin{array}{l}{v}_x^2=v_{x0}^2+2ax\\ 0={\left(35\text{m}/\text{s}\right)}^2+2\left(-5\text{m}/{\text{s}}^2\right)x\\ x=122.5\text{m}\end{array}$

The distance travelled by the car for the reaction time $t$ is calculated as,

  $\begin{array}{c}{x}_t=v_{x0}t\\ =\left(35\text{m}/\text{s}\right)\left(0.500\text{s}\right)\\ =17.5\text{m}\end{array}$

The total distance travelled by the car is calculated as,

  $\begin{array}{c}{x}_t+x=17.5\text{m}+122.5\text{m}\\ =14\text{0}\text{m}\end{array}$

The distance traveled by the car is calculated as,

  $\begin{array}{c}x=v_{x0}t+\frac{1}{2}a{t}^2\\ =\left(35\text{m}/\text{s}\right)\left(2.5\text{s}\right)+\frac{1}{2}\left(3\text{m}/{\text{s}}^2\right){\left(2.5\text{s}\right)}^2\\ =96.875\text{m}\end{array}$

The total distance required to travel to make it through the light is calculated as,

  $\begin{array}{c}{x}_t=96.875\text{m}+17.5\text{m}\\ =\text{114}\text{.375}\text{m}\end{array}$

Conclusion:

Therefore, the distance needed by the car to stop safely is $14\text{0}\text{m}$ and maximum distance between the car and the light so that the car can make it through the light before it turns red is $96.875\text{m}$ .