# Homework Help: Basic Kinematics Question

1. Aug 30, 2009

### nhsu

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

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?

2. Relevant equations

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

3. 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

2. Aug 30, 2009

### 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.