physics
posted by nikki .
A car travelling at a constant speed of 136 km/hr passes a tropper hidden behind a billboard. One second after the speeding car passes the billboard, the tropper sets in a chase after the car wit ha constant acceleration of 3.8 m/s^2. How far does the trooper travel before he overtakes the speeding car?

The distance traveled is the same.
distancespeeder: velocity*time change 136km/hr to m/s
distancetrooper: 1/2 3.8 (t1)^2
set them equal, solve for t. then, go back and solve for distance in either equation 
The answer above is WRONG.
136 km/h = 38.3333 m/s
x = x0 + vi*t + .5*a*t^2
xtropper = 0 + 0 + .5*3.8*t^2
xspeeder = 0 + 38.3333t + 0
xtropper = xspeeder
.5*3.8*t^2=38.3333t
t = 20.17542105
xtropper = .5*3.8*20.17542105^2
xtropper = 773.3904676