Physicss
posted by Raf on .
1. Carlos jogs in a straight line at a constant speed of 1.5 m/s. He passes by Victoria, who, 10 seconds after Carlos had passed her, starts accelerating at a constant rate of 0.50 m/s^2
a) How much time after the passing of Carlos does it take Victoria to catch up to him?
b) What distance did they travel?

Assuming that Victoria was standing still when Carlos passed her, we have the following equations, calling t=0 when he passes her:
Carlos: s = 1.5t
Victoria: s = .25(t10)^2
set them equal to find when they have traveled the same distance  that is, when she passes him.
1.5t = .25(t^2  20t + 100)
1.5t = t^2/4  5t + 25
t^2/4  6.5t + 25 = 0
t^2  26t + 100 = 0
t = 13 ± √69
t = 4.7, 21.3
Since Victoria does not start till t=10, the only answer is 21.3, or 11.3 seconds after she started running.
Plug in times to get distances.