Challenging police car question

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In summary: Where is the cop at 3.75 sec.?What should you do after answering these questions?In summary, the problem involves a policeman accelerating at a constant rate of 2.5 m/s^2 to catch a speeder who is initially traveling at 40 m/s. The speeder decelerates at a constant rate of 4 m/s^2 until he reaches the speed limit of 25 m/s. To solve the problem, one must determine the times and distances traveled by both the policeman and the speeder. The equations Vf=Vi+at, x=Vit+.5at^2, and Vf^2=Vi^2+2ax can be used to solve the problem. After solving
  • #1
jayjay12077
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Homework Statement


A policeman sitting on a side of the road sees a speeder drive by him at 40 m/s. At the same time the speeder passing the policeman 2 things happen:
1. the policeman steps on the gas and begins to accelerate fowards at a rate of 2.5 m/s^2 until he reaches 35m/s
2. the speeder hits the breaks and begins to slow down at a rate of 4 m/s^2 until he reaches the speed limit of 25m/s

A) How long will it take for the cop to catch the speeder?
B) How far will the policeman have to travel before he reaches the speeder


Homework Equations


Vf=Vi+at
x=Vi(t)+(.5)at^2
Vf^2=Vi^2+2ax


The Attempt at a Solution


I tried to solve it by setting the distance that the speeder goes equal to that of the cop and just plugging in initial velocity, acceleration, and time for both, but I did not get the right answer
 
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  • #2
I think the problem with that approach is that the speeder and the cop don't accelerate for equal amounts of time. Speeder decelerates until he gets to 25 m/s from 40 m/s, while the cop accelerates until he gets to 35 m/s from 0 m/s.
 
  • #3
jayjay12077 said:

Homework Statement


A policeman sitting on a side of the road sees a speeder drive by him at 40 m/s. At the same time the speeder passing the policeman 2 things happen:
1. the policeman steps on the gas and begins to accelerate forwards at a rate of 2.5 m/s^2 until he reaches 35m/s
2. the speeder hits the breaks and begins to slow down at a rate of 4 m/s^2 until he reaches the speed limit of 25m/s

A) How long will it take for the cop to catch the speeder?
B) How far will the policeman have to travel before he reaches the speeder


Homework Equations


Vf=Vi+at
x=Vi(t)+(.5)at^2
Vf^2=Vi^2+2ax


The Attempt at a Solution


I tried to solve it by setting the distance that the speeder goes equal to that of the cop and just plugging in initial velocity, acceleration, and time for both, but I did not get the right answer
Hello jayjay12077. Welcome to PF !

Exactly what did you plug into what?

What were your results?

How long does it take the patrolman to reach 35 m/s? How far does he travel in this time?

How long does it take the speeder to slow to 25 m/s? How far does he travel in this time?

What should you do after answering these questions?
 
  • #4
SammyS. Well i got that the cop would take the cop 14 sec. to reach his final velocity. In that time he travels 245 meters. The speeder takes 3.75 seconds to slow down to his final velocty and during that time he travels 121.875 meters. So i thought that since they each had to travel the same distance that I use the equation x=Vit+.5at^2 for the cop and then for the speeder and set the two equations equal to each other. But when I did that and simplifed I got 35t=25t and that does not make sense
 
  • #5
The values for time and displacements for the accelerated portions of the motion seem to be OK. The last part does not seem OK. How did you get this 25t=35t?
 
  • #6
jayjay12077 said:
SammyS. Well i got that the cop would take the cop 14 sec. to reach his final velocity. In that time he travels 245 meters. The speeder takes 3.75 seconds to slow down to his final velocity and during that time he travels 121.875 meters. So i thought that since they each had to travel the same distance that I use the equation x=Vit+.5at^2 for the cop and then for the speeder and set the two equations equal to each other. But when I did that and simplified I got 35t=25t and that does not make sense
After 14 sec., the cop has a constant velocity (zero acceleration).

After 3.75 sec., the speeder has a constant velocity (zero acceleration).

Where is the speeder at 14 sec.?
 

What is the "Challenging Police Car Question"?

The "Challenging Police Car Question" is a thought experiment that poses the dilemma of whether a person should challenge a police car that is speeding and breaking traffic laws in order to prevent potential harm to others.

Why is the "Challenging Police Car Question" important?

This question raises important ethical and moral considerations about the role and responsibility of law enforcement in society, and the potential consequences of challenging authority.

What are the arguments for challenging a police car?

Some argue that it is the duty of citizens to hold law enforcement accountable for their actions and to prevent potential harm to others, regardless of the consequences. Others argue that challenging a police car could lead to dangerous and potentially deadly confrontations.

What are the arguments against challenging a police car?

Opponents argue that challenging a police car could put oneself in danger and could potentially escalate the situation. They also argue that law enforcement officers are trained professionals and should be trusted to make decisions in emergency situations.

How does this question relate to real-life situations?

This question reflects the complex and sometimes conflicting dynamics between citizens and law enforcement, and the difficult decisions that individuals may face in challenging authority for the greater good.

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