Calculating Time and Distance for a Patrol Car Chasing a Speeder

  • Thread starter Thread starter MatthewHaas
  • Start date Start date
  • Tags Tags
    Car
AI Thread Summary
A speeder traveling at 121 km/h is pursued by a patrol car that accelerates from rest at 8.3 km/h/s until it reaches 201 km/h. The patrol car takes approximately 30.4 seconds to catch the speeder, with both vehicles traveling about 1.9 km during the chase. The calculations involved converting speeds to meters per second and using kinematic equations to determine the time and distance. The discussion highlights the importance of careful problem-solving and the potential for errors if not approached methodically. Practice and clear organization of calculations are emphasized as key to mastering similar physics problems.
MatthewHaas
Messages
15
Reaction score
0
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!
 
Physics news on Phys.org
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.
 
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?
 
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!
 
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top