Consider the street pattern shown in Fig. 2–46. Each inter-section has a traffic signal, and the speed limit is 40 km/h. Suppose you are driving from the west at the speed limit. When you are 10.0 m from the first intersection, all the lights turn green. The lights are green for 13.0 s each. (a) Calculate the time needed to reach the third stoplight. Can you make it through all three lights without stopping? (b) Another car was stopped at the first light when all the lights turned green. It can accelerate at the rate of$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$to the speed limit. Can the second car make it through all three lights without stopping? By how many seconds would it make it, or not make it?

FIGURE 2-46Problem 65

You can use the kinematic equations to solve questions related to objects moving with uniform acceleration during their motion.

If “u” is the initial velocity of an object of mass “m”which is moving with uniform acceleration “a” and attains final velocity “v” in time “t” then the kinematics equation of motion connecting these variables can be written as:

$v=u+at$

The speed limit of the rod is:

$\begin{array}{c}v=40\mathrm{km}/\mathrm{h}\\ =\left(40\mathrm{km}/\mathrm{h}\right)\times \left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\times \left(\frac{1\mathrm{h}}{3600s}\right)\\ =11.11\mathrm{m}/\mathrm{s}\end{array}$

You are driving at the speed limit. Therefore, the speed of your car is $v=11.11\mathrm{m}/\mathrm{s}$.

You are at a distance of 10.0 m from the spotlight. According to the figure, in order to reach the third stoplight, you have to cover the following distance:

$\begin{array}{c}x=10+15+50+15+70\\ =160\mathrm{m}\end{array}$

Since you are moving at a constant speed; therefore, the time taken to cover this distance will be:

$\begin{array}{c}\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\\ t=\frac{x}{v}\\ =\frac{160\mathrm{m}}{11.11\mathrm{m}/\mathrm{s}}\\ =14.40\mathrm{s}\end{array}$

Thus, the time needed to reach the third stoplight is 14.40 s. However, lights will remain green for 13.0 s only. Therefore, you cannot pass through all three stoplights without being stopped.

Another car starts its journey from the first stoplight. Thus, it has to travel a distance of 150 m in order to reach the third stoplight.

The initial velocity of the car is $u=0\mathrm{m}/\mathrm{s}$.

The acceleration of the car to reach the speed limit (i.e., $v\text{'}=11.11\mathrm{m}/\mathrm{s}$) is $a=2.00\mathrm{m}/{\mathrm{s}}^2$.

Let $t\text{'}$be the time taken by the car to reach the speed limit from rest.

Using the equation of motion, you will get:

$\begin{array}{c}v\text{'}=u+at\text{'}\\ t\text{'}=\frac{v\text{'}-u}{a}\\ =\frac{\left(11.11-0\right)\mathrm{m}/\mathrm{s}}{2.00\mathrm{m}/{\mathrm{s}}^2 }\\ =5.55\mathrm{s}\end{array}$

Thus,the time needed by another car to reach the speed limit is 5.55 s.

Use the equation of motion to find the distance $s\text{'}$covered by the car to attain the speed limit starting from rest.

$\begin{array}{c}{\left(v\text{'}\right)}^2-u^2=2as\text{'}\\ {\left(11.11\mathrm{m}/\mathrm{s}\right)}^2-{\left(0\mathrm{m}/\mathrm{s}\right)}^2=2\left(2.00\mathrm{m}/\mathrm{s}\right)s\text{'}\\ s\text{'}=30.86\mathrm{m}\end{array}$

Thus, in order to reach the third stoplight, the car has to cover a distance of $150\mathrm{m}-30.86\mathrm{m}=119.14\mathrm{m}$.

If $t\text{'}\text{'}$is the time taken by the other car in order to cover the remaining distance of 119.14 m at a constant speed of $11.11\mathrm{m}/\mathrm{s}$, then:

$\begin{array}{c}t\text{'}\text{'}=\frac{119.14\mathrm{m}}{11.11\mathrm{m}/\mathrm{s}}\\ =10.75\mathrm{s}\end{array}$

Thus, the total time taken by the other car to reach the third stoplight is:

$\begin{array}{c}t\text{'}+t\text{'}\text{'}=\left(5.55+10.75\right) \mathrm{s}\\ =16.3\mathrm{s}\end{array}$

However, lights will only remain green for 13.0 s, and the second car requires 16.3 s to reach the third stoplight. Therefore, the second car cannot pass through all three stoplights without being stopped. It cannot pass the third stoplight by$\mathbf{16}\mathbf{.}\mathbf{3}\mathbf{s}-\mathbf{13}\mathbf{.}\mathbf{0}\mathbf{s}=\mathbf{3}\mathbf{.}\mathbf{3}\mathbf{s}$.