Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

You are driving your car on a very cold late Fall day. You clear a turn and see a couple of pedestrians standing at the cross walk. They are eager to cross the road and to get into the warmth of their apartment as soon as possible. You have two options: continue driving your car as you were without lowering your speed and drive right by the pedestrians OR slow down, stop right at the crosswalk, and yield to the pedestrians. Although by Virginia law the choice is clear, what about Physics laws? Which scenario (passing by or slowing down and stopping at the crosswalk to yield) will minimize the time the pedestrians are out in the cold freezing before they can cross the road?
Make the following assumptions in your argument. Before you noticed the pedestrians, you are moving with a constant velocity v=22 miles/hour. The distance at which you noticed the pedestrians is D=23 meters. Write down a symbolic expression for the amount of time, tpass , the pedestrians will have to wait till they cross the road if you simply drive by without slowing down or speeding up.
Write down a symbolic expression for the amount of time, tstop, the pedestrians will have to wait till they cross the road if you slow down, come to a complete stop at the crosswalk and yield to the pedestrians.

Sagot :

hamzaahmeds

Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

[tex]t_{pass} = \frac{D}{v}[/tex]

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]

t_pass = 2.34 m

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]

Now, for the passing time we use first equation of motion:

[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]

t_stop = 4.68 s

Constant velocity is the velocity which covers the same distance for each interval of the time.

The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

What is constant velocity?

Constant velocity is the velocity which covers the same distance for each interval of the time.

It can be given as,

[tex]v=\dfrac{x}{t}[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

According to the third equation of the motion acceleration can be given as,

[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]

Now, use the first equation of motion, to get the required time,

[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]

Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

For more details about equation of motion, refer to the link:

https://brainly.com/question/8898885

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.