Answer to Question #145792 in Physics for Cath

Question

A jogger running at a constant speed of 4.4 m/s passes a friend who is sitting stationary on a bike. Just as the jogger passes, the biker chases at a rate of 2.6 m/s squared. How much time passes before the bike catches up to the jogger?

Answer 1

$$s_1=s_2\\vt=0.5at^2\\2v=at\\2(4.4)=2.6t\\t=3.4\ s$$