Posted by Ty on .
Hi: We have a homework problem that I have no idea even how to set up! please help!
"A man jogs at a speed of 1.3 m/s. His dog waits 1.4s then takes off running at a speed of 2.8 m/s to catch the man. How far will they have to travel before the dog watches up with the man? Answerin units of m"

physics 
DrBob222,
rate of man = 1.3 m/s
t+1.4 s = time man jogs.
distance covered by man is distance = rate x time = 1.3(t+1.4)
rate of dog = 2.8 m/s.
t = time dog runs.
distance covered by dog is distance = rate x time = 2.8t
Both run the same distance so set distance = rate x time equal to each other. That is
distance man runs = distance dog runs. Check my thinking. Check my work. I found a little over 3 meters. Post your work if you get stuck.