An unmarked police car traveling a constant 80 is passed by a speeder traveling 145 .Precisely 2.50 after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.80 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
I am getting t = 22.06, and its wrong :( Could someone help me out please!
-Thanks
2 answers
You need to provide dimensions with your numbers. That could be why you got the wrong answer.
Since Xf for the police is going to be equal to the Xf of the speeder we know that we can write two equations for the speeder and the police and set them equal to each other in order to solve for t.
first we need to convert
Police: v1=80km/hr-->22.22m/s
Speeder: v2=145km/hr-->40.27 m/s
Police: Xf= vt + (0.5)(a)t^2
Xf=22.22*t+(0.5)(1.8)(t-2.5)^2
Speeder:
Xf=40.27*t
Solve for t by setting the eqns equal to each other
So we get:
0.9t^2-22.55t+5.63
use the quad. eqn to solve for t
t=24.8
first we need to convert
Police: v1=80km/hr-->22.22m/s
Speeder: v2=145km/hr-->40.27 m/s
Police: Xf= vt + (0.5)(a)t^2
Xf=22.22*t+(0.5)(1.8)(t-2.5)^2
Speeder:
Xf=40.27*t
Solve for t by setting the eqns equal to each other
So we get:
0.9t^2-22.55t+5.63
use the quad. eqn to solve for t
t=24.8
2 answers