Calculating Time and Distance for a Patrol Car Chasing a Speeder

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In summary, using the given equations and conversion of units, it can be determined that the patrol car will take 30.4 seconds to catch up with the speeder after passing the billboard. During this time, the patrol car will travel 1916.00 m and the speeder will travel 1915.96 m. It is important to take time to write clear and organized solutions to avoid errors.
  • #1
MatthewHaas
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A speeder traveling at a constant speed of 121 km/h races past a billboard. A patrol car pursues from rest with constant acceleration of (8.3 km/h)/s until it reaches its maximum speed of 201 km/h, which it maintains until it catches up with the speeder.


(a) How long does it take the patrol car to catch the speeder if it starts moving just as the speeder passes? (in s)


(b) How far does each car travel? (in km)


Here are the equations I have been using:
Vx=V_ox + (a_x)(t)
X=vt+(1/2)at^2
and the special case...
X-x_o=vt+(1/2)at^2


I first converted 121 km/h to 33.61 m/s, 201 km/h to 55.83 m/s and acceleration of 8.3 km/h/s to 2.3056 m/s^2

Using Vx=v_ox+(a_x)(t) I found the time it took to reach a velocity of 55.83 m/s (since v_ox is 0 m/s at acceleration is 2.3056 m/s^2) to be 24.21 s.

In 24.21 s, the patrol car (car 2) has traveled 675.96 m. (using the distance formula)..(1/2)(2.3056)(24.21)^2

In that same time, car 1(the speeder) has traveled 813.6981 m. (33.61 m/s * 24.21 s)

Next, I used the special case X-x_o=vt+(1/2)at^2

Car 1 x-813.6981 m = 33.61 m/s *t
Car 2 x-675.96 m = (1/2)(2.3056 m/s^2)*t^2

X=33.61t + 813.6981
X=1.1528t^2+675.96

Since X is common, I can set these two equations equal to one another.

1.1528t^2-33.61t-137.738=0

t can now be found using the quadratic equation

a=1.1528, b=-33.61, c=-137.738

I get two answers: 32.797, -3.642 (the second is of course nonsensical)

When I plug back into special case eqs (X=33.61t+813.6981 and 1.1528t^2+675.96) I get 1916.00 m and 1915.96 m which = (roughly) 1.9 km.

33s until the officer reaches the speeder and it takes him 1.9 km to do this.

I believe I have the correct work, but might have a rounding error? The answer in the back of the book (which has similar numbers) is close to what I have as an answer...But when I tried my method using the book's numbers, I didn't get the same answer (although I was close).

Help please?

THANK YOU!
 
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  • #2
I ran it through and agree exactly with your calcs for the accelerated part (first 24.21 seconds). After that it seems to me the acceleration of both cars is zero so good old d = vt does the job. I considered only the separation distance and the speed difference:
t = d/v = (813.698 - 675.96)/(55.83 - 33.61) = 6.19 seconds.
Total time is 24.21 + 6.19 = 30.4 s.
 
  • #3
Thank you, Delphi51. That was very helpful!

And painfully obvious now that I look at it. Any tips on how to "sense" these things? Or just practice?
 
  • #4
Welcome to the forum!

Yeah, it's pretty much all practice. You'll start sensing the fastest and easiest ways to do these problems too. For example, another problem would be if the acceleration wasn't uniform - but increasing too. You can't just use that formula to solve it. To solve a problem like that, you would need to use integration (calculus).

The formula d = Vi * t + 1/2 * a * t^2 is just the integral of a*t with respect to t. When you start taking calculus classes, you'll see this little anomaly.

Cheers!
 
  • #5
Most welcome, Matthew.
Many errors of that kind can be avoided by taking just a little more time when writing the solution. I learned this very quickly after my B.Sc. in physics when I found myself in an unfamiliar situation, learning to teach high school physics. My prof gave a grade 12 assignment and asked us to write it up as we would expect a good grade 12 student to do. I had so much confidence I dashed off that assignment very quickly, and I made several mistakes. And didn't write clear solutions. Really got dinged, and really deserved it. I changed my ways instantly. Now when I see a problem like yours, I immediately make two headings for the two parts of the problem. Maybe a line between the two parts as well. That makes me be aware of what quantities can be carried across the line to the other part.

Good luck!
 

FAQ: Calculating Time and Distance for a Patrol Car Chasing a Speeder

What is a patrol car chasing a speeder?

A patrol car chasing a speeder is a common scenario in law enforcement where a police officer uses their vehicle to pursue a driver who is breaking traffic laws or attempting to flee from the police.

Why do patrol cars chase speeders?

Patrol cars chase speeders in order to enforce traffic laws and maintain public safety. Speeding and other traffic violations can be dangerous and put other drivers at risk, so it is important for police to stop and ticket these individuals.

What happens during a patrol car chase?

During a patrol car chase, the police officer will use sirens and lights to signal the driver to pull over. The officer will then follow the vehicle while maintaining a safe distance, and may use communication with other officers to coordinate the pursuit. The driver will eventually be pulled over and issued a citation or arrested if necessary.

Is it dangerous for a patrol car to chase a speeder?

While patrol car chases can be dangerous, police officers are trained to handle these situations and take precautions to ensure the safety of all individuals involved. However, it is important for drivers to follow traffic laws and not engage in high-speed chases to avoid putting themselves and others at risk.

What should I do if I am being chased by a patrol car?

If you are being chased by a patrol car, it is important to remain calm and pull over when it is safe to do so. Follow the instructions of the police officer and cooperate with their commands. If you believe you are being falsely pursued, you can address the issue in court rather than attempting to flee from the police.

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