How Far Does the Cop Travel to Catch the Speeder?

[nhsu]

Homework Statement

A speeder passes a hiding speed cop at 55m/s. The speed cop puts down his coffee and starts his engine (which takes 5 sec). He then accelerates from rest at 5m/s^2 until he catches the speeder.
How far does the cop travel before he catches up and how fast is the cop going?

Homework Equations

V=x/t --> x=vt
Vf^2 = Vi^2 + 2ax

The Attempt at a Solution

I know that the final distances can be set to equal each other but I am confused on what to put for time. If i can solve for time, I can use the time multiplied by 55m/s to find how far they traveled. Also, if the cop takes 5 seconds to start moving, the speeder is already 275m away

Velocity of the cop
Vf^2 = Vi^2 +2ax
Vf^2 = 0^2 + 2(5)(275-0)
Vf = 52.44 m/s

Time of cop
v=x/t
52.44 = 275m / t
t = 5.24 sec

Xspeeder=Xcop
Vspeeder x Tspeeder = Vcop x Tcop
55 m/s x Tspeeder + 275m = 52.44 m/s x 5.24 sec
But when I solve for Tspeeder I get a negative number and time can't be negative.
Any help would be appreciated!
Thank you

 

[rl.bhat]

The distance moved by the speeder in ( t + 5 ) seconds is equal to the distance traveled by cop in t seconds.
Equate them and solve the quadratic to get the time.

 

[tomcue]

for providing the necessary information and equations to solve this problem. It seems like you have made a good attempt at finding the solution, but there are a few errors in your calculations.

First, let's clarify the given information. The speed cop takes 5 seconds to start his engine, but it is not specified how long it takes for him to accelerate to 5m/s^2. For simplicity, let's assume that it takes him an additional 5 seconds to reach this acceleration. Therefore, the total time it takes for the cop to catch the speeder is 10 seconds.

Next, let's define our variables. We know that the speeder's initial velocity (Vi) is 55m/s and the cop's initial velocity is 0m/s. We also know that the speeder's final velocity (Vf) is 55m/s (since it is traveling at a constant speed) and the cop's final velocity (Vf) is what we are trying to find. The acceleration (a) for both the cop and the speeder is 5m/s^2.

Using the equation Vf^2 = Vi^2 + 2ax, we can solve for the cop's final velocity:

Vf^2 = 0^2 + 2(5)(275-0)
Vf^2 = 2750
Vf = √2750
Vf = 52.5 m/s

Now, to find the distance traveled by the cop, we can use the equation x = vt. We know that the cop's final velocity is 52.5 m/s and the time is 10 seconds. Therefore:

x = (52.5 m/s)(10 sec)
x = 525 m

So the cop travels 525 meters before catching up to the speeder. This makes sense because the speeder has a 275 meter head start and the cop has to travel an additional 250 meters to catch up.

I hope this helps clear up any confusion and good luck with your homework!