A bicyclist is finishing his repair of a flat tire when a friendrides by with a

Question

A bicyclist is finishing his repair of a flat tire when a friendrides by with a constant speed of 3.8m/s. Two seconds later the bicyclist hops on his bike andaccelerates at 2.6 m/s2until he catches his friend. (a) How much time does it take until he catcheshis friend?
1 s
(b) How far has he traveled in this time?
2 m
(c) What is his speed when he catches up?
3 m/s (a) How much time does it take until he catcheshis friend?
1 s
(b) How far has he traveled in this time?
2 m
(c) What is his speed when he catches up?
3 m/s

Explanation / Answer

The position x1, of the friend obeys the equation x1=vt The position x2 of the bicyclist obeys the equation x2=1/2at^2 we want to know the time when the two positions are equal so, x1=x2 vt=1/2a^t2             solve for t t=2v/a=3.273seconds the both travel the same distance so d=vt=(3.273*3.6) d=11.78 c)v=at v=(2.2)*3.273=7.2m/s

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