The problem involves a scenario where a speeder passes a stationary police car, which then accelerates to catch up. The speeder travels at a constant speed, while the police car accelerates uniformly. Participants are tasked with determining the time it takes for the police car to overtake the speeder.
Some participants have attempted calculations and shared their results, while others are exploring different methods to arrive at the solution. There is an ongoing dialogue about the correct equations to use and the assumptions involved in the problem.
Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring that both vehicles are at the same position at the same time for the overtaking to occur.
bigzee20 said:Homework Statement
A speeder passes a parked police car at a constant speed of 27.6 m/s. At the instant, the police car starts from rest with a uniform acceleration of 2.27 m/s^2. How much t passes before the speeder is overtaken by the police car.
I tried solving for t using t=Vf-Vi /a and i get 12.15sec, which is wrong what am i doing wrong?
bigzee20 said:OMMMMG I love u I GOT IT!
x=1/2a(t)^2 + Vi(t)
Cops= 1/2(2.27)(12.15)+0(12.15) = 13.179025
Speeder= 1/2(0)12.15+27.6(12.15) = 335.34
335.34/13.179025 = 24.32s