Question #145792
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?
1
Expert's answer
2020-11-23T10:26:55-0500
s1=s2vt=0.5at22v=at2(4.4)=2.6tt=3.4 ss_1=s_2\\vt=0.5at^2\\2v=at\\2(4.4)=2.6t\\t=3.4\ s


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