Question

A person driving her car at $\mathbf{35}\mathbf{km}\mathbf{/}\mathbf{h}$approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection (Fig. 2–49). Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car’s maximum deceleration$-\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$is whereas it can accelerate from$\mathbf{45}\mathbf{km}\mathbf{/}\mathbf{h}$to$\mathbf{65}\mathbf{km}\mathbf{/}\mathbf{h}$in 6.0 s. Ignore the length of her car and her reaction time.

FIGURE 2-49 Problem 73.

Short answer

She would travel $26.852\mathrm{m}$before the light turns red and try to stop the car to be a legal driver.

Step-by-step solution

Step 1. Significance of the moving or stoppage of the car when depending upon the traffic signal lights

The indications of the traffic lights are as follows.

• At the green signal, the car moves in the required direction.
• At the red signal, the car stops its motion.
• At the yellow signal, the driver is warned that the traffic signal is going to be red.

The equations of motions come to play in analyzing the car'smotion near the traffic signal lights.

With the help of these equations, the required results can be obtained.

Step 2. Identification of the given data in both cases

Case (I). The stoppage of the car.

• The deceleration of the car is$a=-5.8\mathrm{m}/{\mathrm{s}}^2$.
• The final velocity of the car is $v_f=0\mathrm{m}/\mathrm{s}$.
• The initial velocity of the car is $v_i=35\mathrm{km}/\mathrm{h}\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{hr}}\right)=9.722\mathrm{m}/\mathrm{s}$.

Case (II). Boosting the car’s speed through the intersection.

• The first initial velocity of the car is $v_i=45\mathrm{km}/\mathrm{h}\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{hr}}\right)=12.5\mathrm{m}/\mathrm{s}$.
• The second initial velocity of the car is

$v_f=65\mathrm{km}/\mathrm{h}\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{hr}}\right)=18.055\mathrm{m}/\mathrm{s}$.

• The time taken by the car is $\Delta t=6\mathrm{s}$.

Step 3. Determination of the distance traveled by the car in case (I)

The distance moved by the car in the case of the stoppage of the car.

$v_f^2=v_i^2+2a\Delta s_i$

Substitute the values 0 m/s for $v_f$, $9.722\mathrm{m}/\mathrm{s}$ for $v_i$, and $-5.8\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{c}{\left(0\mathrm{m}/\mathrm{s}\right)}^2={\left(9.722\mathrm{m}/\mathrm{s}\right)}^2+2\times \left(-5.8\mathrm{m}/{\mathrm{s}}^2\right)\times \Delta s_i\\ \Delta s_i=8.148\mathrm{m}\end{array}$

So, the car was stopped by her on time.

Step 4. Determination of the acceleration of the car in case (II)

The acceleration of the car can be evaluated using the first equation of motion. The equation of motion can be expressed as.

$\begin{array}{c}{v}_f=v_i+a_1\Delta t\\ a_1=\frac{v_f-v_i}{\Delta t}\end{array}$

Substitute the values 18.055 m/s for $v_f$, $12.5\mathrm{m}/\mathrm{s}$for $v_i$, and $6\mathrm{s}$for $\Delta t$in the above equation.

$\begin{array}{c}{a}_1=\frac{18.055\mathrm{m}/\mathrm{s}-12.5\mathrm{m}/\mathrm{s}}{6\mathrm{s}}\\ =0.926\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 5. Determination of the distance travelled by the car in 2 s in case (II)

The distance moved by the car in 2 s can be expressed as.

$\Delta s_f=v_{i}t+\frac{1}{2}{a}_1{t}^2$

Here, t is the time taken by the car whose value is 2 s.

Substitute the values as 2 s for t, $0.926\mathrm{m}/{\mathrm{s}}^2$for $a_1$, and $12.5\mathrm{m}/\mathrm{s}$for $v_i$in the above equation.

$\begin{array}{c}\Delta s_f=12.5\mathrm{m}/\mathrm{s}\times \left(2\mathrm{s}\right)+\frac{1}{2}\times 0.926\mathrm{m}/{\mathrm{s}}^2\times {\left(2\mathrm{s}\right)}^2\\ =26.852\mathrm{m}\end{array}$

The person is 28 m away from the intersection, and the intersection is 15 m wide.

The total distance can be given as.

$\begin{array}{c}L=28\mathrm{m}+15\mathrm{m}\\ =43\mathrm{m}\end{array}$

The calculated distance of the car in 2 s is less than the total length. This implies that the distance is not sufficient to safely pass the given intersection.

Hence, the car should be stopped by the person.

She should travel $26.852\mathrm{m}$before the light turns red and try to stop the car to become a legal driver.