# Cop's Accel and speeder's velocity?

1. Sep 22, 2005

### stealthhatch

i have this problem in physics i need to get done. i've tried numerous different paths to get the right answer but i just can't think it through.

the problem is:
A speeder passed a parked police car at a constant speed of 35.5 meters/sec.. At that instant the police car starts from rest with a uniform aceleration of 2.29 meters/sec. How much time(sec.) passes before the speeder is overtaken by the police car?

I've spent literally 3hrs. tring to figure out this problem, but i just can't think it through, and my professor accidently only solved for the time it would take for the police car reach the same velocity as the car.

So far i have calculated:
-that it takes 15.5021834 sec for the police car to reach the speed of the speeding car and begin gaining on it.

-At that point the speeding car has traveled 550.3275109 meters from the point where he passed the cop.

- the police car travels a distance of 275.163755 meters in order to get to 35.5 meters/sec.

- and at that point where the police car is traveling 35.5 m/sec the two cars are 275.1637554 meters apart.

after a long time of thinking this was the only way i thought of to get the solution, but it seems like a dead end.
i just can't think of how this information would be used to get the total time for the two to reach the same point at the same time.
can someone point me in the right direction on how to get this answer?

2. Sep 22, 2005

### stunner5000pt

so for both of the them the TIME is going to be the same

Also visualize - when the cop car catches the speeder what else is equal??

so we know that $t_{cop car} = t_{speeder}$
now write each time relating what you know (velocity, acceleration) and that COMMON thing(even if you dont know it, use it, because solving for it, will help you solve for hte time)

Last edited: Sep 22, 2005
3. Sep 22, 2005

### stealthhatch

i don't really get what you mean. i know the time and distance are equal for both of the cars, but every EQ. seems to require info not given in the question, and i also don't know what to use for acceleration for the car with constant velocity, if anyone could give an example of how to solve a similar problem it would be much appreciated.

4. Sep 22, 2005

### stunner5000pt

the distance the cop covered in so much time is
d = 1/2 at^2 correct?

the distance the speeder covered is d = vt
The distances are equal... good so can you solve for distnace or time using the above equations?

5. Sep 22, 2005

### stealthhatch

thank you,
i don't know why i didn't think of that before( i guess i'm stupid).
the whole time i was trying to equate the formulas together i forgot that the value of them didn't matter as long as they were the same. i guess you said that right away but i just did't get.