1. Apr 15, 2012

knnguyen

1. The problem statement, all variables and given/known data

An unmarked police car traveling a constant 80 km/h is passed by a speeder traveling 125 km/h.
Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s^2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

2. Relevant equations

ΔX (speeder) = V(speeder)T → (125km/h) (1m/s / 3.6km/h) (T) = (34.72222T)m

ΔX (police) = V (police)(3.00s) →(80km/h)(1m/s / 3.6km/h) (3.00s) = 66.666667m
ΔX (police during accelerated motion) = V (t-3.00) + 1/2(2.40m/s^2)(t-3.00)^2
Total ΔX (police) = 66.666667 + 66.666667(t-3.00)+1/2(2.40)(t-3.00)^2

ΔX (speeder) = ΔX (police)

3. The attempt at a solution

4.0
4.1
2.5
0.64

Were all my answers but all are wrong. Please let me know what I did wrong. Thank you so much in advance!

2. Apr 15, 2012

tms

Re: many failed attempts on this displacement/acceleration/velocity problem. please h

You know that your third and fourth answers can't possibly be right, because the police car does not even start to accelerate until three seconds have passed.

Use the basic kinematic equations, and write down an equation for the speeder and another for the cop.