Q40P - Fundamentals Of Physics

Question

You are driving towards a traffic signal when it turns yellow. Your speed limit is the legal speed limit of  ${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{55}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$; your best deceleration rate has a magnitude  $\mathit{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{18}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. Your best reaction time to begin braking is  $\mathit{T}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{ }\mathit{s}$. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at $\mathbf{55}\mathbf{ }\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$  if the distance to the intersection and the duration of yellow light are (a) $\mathbf{40}\mathbf{ }\mathit{m}$  and $\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{ }\mathit{s}$ (b) $\mathbf{32}\mathbf{ }\mathit{m}$  and  $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathit{s}$.Give an answer of brake, continue, either (if either strategy works) or neither (if neither strategy works and yellow duration is inappropriate).

Step-by-Step Solution

Answer

(a). Car should stop in the given distance.

(b). Car should not stop in the given distance and as it will be stuck ahead of the intersection.

Given information

Initial velocity is $55 \text{km/h}$

Acceleration (a) is  $5.18 {\text{m/s}}^{\text{2}}$

Distance $x_1=40 \text{m}$, Time $T_1=2.8 \text{s}$

Distance $x_2=32 \text{m}$, Time  $T_2=1.8 \text{s}$

Understanding the concept

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for the distance and velocity, the distance during reaction time, breaking time and hence total distance can be calculated. Using the calculated quantities it can be determine whether driver should stop or continue to move at $\mathbf{55}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$  at different distances and duration.

Formula:

The velocity in kinematic equation is given by,

 $v_f^2=v_0^2+2ax$

(a) Determination of driving decision for distance 40 m and time 2.8 sec

Let’s consider the direction of motion is +x.

 so  $a=-5.18 {\text{m/s}}^{\text{2}}$  and

 $\begin{array}{rcl}{v}_0& =& 55 \text{km/h}\\ & =& 15.28 \text{m/s}\end{array}$

The distance traveled during the reaction time is 

 

 $\begin{array}{rcl}{d}_r& =& v_{0}T\\ d_r& =& \left(15.28 \frac{\text{m}}{\text{s}}\right)\left(0.75\right)\\ d_r& =& 11.46 \text{m}\end{array}$

The distance traveled during the breaking is 

 

$\begin{array}{rcl}{v}^2& =& {v_0}^2+2a{d}_b\\ d_b& =& -\frac{15.28 {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{2\left(-5.18 {\frac{\text{m}}{{\text{s}}^{\text{2}}}}^2\right)}\\ & =& 22.53 \text{m}\end{array}$ 

Thus, total distance traveled is

 $\begin{array}{rcl}{d}_r+d_b& =& 11.46 \text{m}+22.53 \text{m}\\ & =& 34.0\text{m}\end{array}$

So driver is able to stop in time. If he continues at $v_0$, the car would enter the intersection

 $\begin{array}{rcl}t& =& \frac{40 \text{m}}{5.28 \text{m/s}}\\ & =& 2.6\text{s}\end{array}$

This is sufficient to enter the intersection before light turns. But it is too small to continue driving. So it would be advisable for the driver to stop.

 

(b) Determination of driving decision for distance 32 m and time 1.8 sec

Total distance to stop is greater than the distance to the intersection; hence the driver can’t stop without the front of the car being few meters into the consideration. 

Time to reach this distance is  $\frac{32}{15.28} \text{m/s}=2.1 \text{s}$.

This time is greater than that of the available time. So this is tricky situation. Driver would not be able to stop and would be able to just cross the intersection.