Can someone help with a physics problem?

Question

A car is traveling at a constant speed of 30 m/s passes a highway patrol police car which is at rest. The police officer accelerates at a constant rate of 3.0 m/s^2 and maintains this rate of acceleration until he pulls next to the speeding car. Assume that the polic car starts to move at the moment the speeder passes his car. Determine the (a) time required for the police officer to catch the speeder and (b) distance traveled during the chase.

I tried using the equation deltat = (Vf-Vi)/a but, apparently, it's wrong...>.<

Answer 1

the speeder maintains constant speed throughout the time t it takes for the police car to intercept. Call d the distance traveled by both cars to the intercept point. THen
d=30*t for the speeder
and
d=.5*3*t^2 for the police car
since the distance to intercept is the same
30*t=.5*3*t^2
divide by 3*t and multiply by 2
20=t

the distance is found by plugging into either equation for d
30*20=600
.5*3*400=600

j

Answer 2

To solve this problem you would need to write a position equation for each car:

X = Xo + Vo*t + 1/2*a*t^2

Speeder:
X = 0 + 30*t

Cop:
X = 0 + 1/2*3*t^2

Since the problem ends when the cop and speeder are at the same position you can set these two equations equal to eachother and solve for t which gives you 20 seconds. To then figure out how far they traveled during the chase you would just plug the 20 seconds back into either of the two origianal equations and get 600m