quite the mind boggling question

[Krz]

"An unmarked police car, traveling a constant 90km/h, is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the police man steps on the accelerator. If the police car's acceleration is 2.00 m/s^2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at a constant speed)?"

 

blew my mind away

[nanogrinder]

i got 14.8

 

first i took 90km/h and 140km/h then converted them to m/s which comes out to be 25m/s and 38.9m/s respectively. we will call the car 25,x and car 38.9, z

 

after this i use the equation

 

d=Vot+.5at²

 

where d=distance, Vo=original velocity, a=acceleration, t=time

 

the first second for car x travels 25m, using d=25(1)+.5(0)(1²) the last part is zeroed out because there is no acceleration. car z travels 38.9m, d=38.9(1). we need to now make two equations involving car x and z that are set equal to each other,but we must also add in those other two answers in them.

 

since they are going to be at an equal distance in a certain amount of time, we can do this

 

38.9t+38.9= 25+25t+.5(2)t² (btw we had to add the other values 38.9 and 25 into the equation) now get everything to one side -t²+13.9t+13.9=0

now just put the numbers into the quadratic formula then volia 14.8s

 

i hoped that help explain, its hard to get these kinds of points across on the net.

[swansont]

The point of the HW forum is not to do the problem for the poster, but to help him/her solve the problem and learn how to do it.

[Krz]

haha its cool, i can learn from seeing what he did, although his answer was off by 1, i dont know :S

[nanogrinder]

The point of the HW forum is not to do the problem for the poster, but to help him/her solve the problem and learn how to do it.

 

sorry about that.

 

p.s. krz i forgot to add in the second it took for the officer to react.

[Krz]

cool cool, thanks