Chapter 2, Problem 32P

To determine

To identify: Whether a person will try to stop or speed up to cross the intersection before the light turns red.

Answer to Problem 32P

A person should stop the car.

Explanation of Solution

Given:

Initial speed of car, $u=45\text{ km/h}$

Length, $L_1=28\text{ m}$

  $L_2=15\text{ m}$

Formula used:

  $v^2=u^2+2as$

Where, $v$ = final velocity

  $u$ = initial velocity

  $s$ =distance

  $a$ = acceleration

Calculation:

The initial speed of car in m/s is

  $\begin{array}{c}45\text{km}/\text{h}=45\times \frac{5}{18}\\ =\frac{5\times 5}{2}\\ =12.5\text{ m/s}\end{array}$ .

The maximum distance covered by her if she decided to stop,

  $\begin{array}{l}{v}^2=u^2+2as\\ 0={\left(12.5\right)}^2+2\left(-5.8\right)x_1\\ x_1=\frac{{\left(12.5\right)}^2}{2\left(-5.8\right)}\\ x_1=13.5\text{ m}\end{array}$

The distance $x_1$ is less than $L_1$ so she can safely stop.

Now, if she decides to accelerates from $12.5\text{ m/s}$ to $18.1\text{ m/s}$ in $6\text{ s}$ then acceleration of car is,

  $\begin{array}{l}a=\frac{v-u}{t}\\ a=\frac{18.1-12.5}{6}\\ a=0.93{\text{ m/s}}^2\end{array}$

The yellow light is available only for two seconds then after red light would turn on.

Therefore, the distance covered in $2\text{ s}$ is,

  $\begin{array}{l}s=ut+\frac{a{t}^2}{2}\\ x_2=\left(12.5\times 2\right)+\frac{0.93{\left(2\right)}^2}{2}\\ x_2=25+1.86\\ x_2=26.86\text{ m}\end{array}$

The distance $x_2$ is less than $L_1$ . This means a person will not safely pass the intersection.

Conclusion:

Therefore, a person should stop the car.