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
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
To solve this problem, we can start by setting up equations of motion for both the speeder and the police car. The speeder is initially traveling at a constant speed of 145 m/s, so its equation of motion is:
s1 = 145t
Where s1 represents the distance traveled by the speeder and t represents time.
The police car, on the other hand, starts from rest and undergoes constant acceleration. Its equation of motion can be written as:
s2 = 0 + (1/2)a(t^2)
Where s2 represents the distance traveled by the police car and a represents its acceleration.
Since the police car starts moving 2.50 seconds after the speeder passes, we can substitute (t + 2.50) for t in the equation of motion for the police car. This will give us the distance traveled by the police car at the same time as the speeder passes:
s2 = 0 + (1/2)a((t + 2.50)^2)
To find the time at which the police car overtakes the speeder, we need to set s1 equal to s2 and solve for t:
145t = (1/2)a((t + 2.50)^2)
Now, we can substitute the given values:
145t = (1/2)(1.80)((t + 2.50)^2)
Simplifying:
145t = 0.90(t^2 + 5t + 6.25)
145t = 0.90t^2 + 4.5t + 5.625
0.90t^2 - 140.5t + 5.625 = 0
Now we can solve this quadratic equation for t using any preferred method such as factoring, completing the square, or using the quadratic formula.