Chapter 2, Problem 76GP

(a)

To determine

The time needed to reach the third stoplight.

To identify: Whether the car behind the first stoplight make it through all three lights without stopping.

Answer to Problem 76GP

The time needed to reach the third stoplight is $11.52\sec$ . Yes, the car behind the first stoplight can make it through all three lights without stopping.

Explanation of Solution

Given:

The given situation is shown below.

  

The speed of the car behind the first stoplight is $50\text{km/hr}=\frac{\text{50×1000m}}{\text{3600s}}=13.89\text{m/s}$ . The total distance needed to travel and cross the third stoplight is $\text{10m}\text{+15m}\text{+}\text{50m +15m}\text{+}\text{70m}\text{=160m}$ . The lights are green for $13\sec$ each.

Formula used:

The time is given by the formula

  $t=\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Calculation:

The time needed to cross the third stoplight is

  $\begin{array}{c}t=\frac{\text{160m}}{\text{13}\text{.89m/s}}\\ =\text{11}\text{.52}\sec \end{array}$

The time needed to reach the third stoplight is $11.52\sec$ . Hence, yes, the car behind the first stoplight can make it through all three lights without stopping when all lights are green.

(b)

To determine

To identify: Whether the second car stopped at the first stoplight make it through all three lights without stopping.

Answer to Problem 76GP

No, the car cannot make it through all the three stoplights without stopping.

Explanation of Solution

Given:

The given situation is shown below.

  

The speed of the car behind the first stoplight is $50\text{km/hr}=\frac{\text{50×1000m}}{\text{3600s}}=13.89\text{m/s}$ . The total distance needed to travel and cross the third stoplight is $\text{10m}\text{+15m}\text{+}\text{50m +15m}\text{+}\text{70m}\text{=160m}$ . The lights are green for $13\sec$ each.

Formula used:

Newton’s first and second equation of motion is

  $\begin{array}{l}v=u+at\\ s=ut+\frac{1}{2}a{t}^2\end{array}$

Calculation:

Using Newton’s first equation of motion, the duration of the acceleration is

  $\begin{array}{l}v=u+at\\ t=\frac{v-u}{a}=\frac{\text{13}\text{.89m/s}-\text{0m/s}}{{\text{2m/s}}^{\text{2}}}\\ t_{acc}=6.94\sec \end{array}$

Now, using this time duration in Newton’s second equation of motion to find the distance traveled by car is

  $\begin{array}{l}s=ut+\frac{1}{2}a{t}^2\\ s=0+\frac{1}{2}\times 2m/s^2\times {(}^{6.94}\\ s=48.2m\end{array}$

The total time until all lights will be green is $13 \sec$ . The remaining time till the car will move with its final velocity $v=13.89 \text{m/s}$ is $13-6.94=6.06 \sec$ . Thus, the distance covered in the remaining time with the constant speed is

  $\begin{array}{c}{d}_{cons\tan t speed}=v\times t\\ \text{=13}\text{.89m/s×6}\text{.06s}\\ d_{cons\tan t speed}=84.2\text{m}\end{array}$

The total distance covered by the car is $48.2\text{m}+84.2\text{m}=132.4\text{m}$ which is not enough.

Conclusion:

Hence, the car cannot make it through all the three stoplights without stopping.