# Patrol car chasing a speeder

1. Sep 15, 2010

### MatthewHaas

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).

THANK YOU!

2. Sep 15, 2010

### Delphi51

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. Sep 16, 2010

### MatthewHaas

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. Sep 16, 2010

### thehacker3

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. Sep 16, 2010

### Delphi51

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!