- #1

stealthhatch

- 3

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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?