SOLUTION: A stationary cyclist is passed by another cyclist riding with a constant speed of 3.4 m/s. Immediately as the other bike passes the stationary bicyclist hops on his bike and accele

Question 518946: A stationary cyclist is passed by another cyclist riding with a constant speed of 3.4 m/s. Immediately as the other bike passes the stationary bicyclist hops on his bike and accelerates at 2.4m/s2 until he catches up. (a) How much time does it take to catch up and how far has he travelled in this time?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A stationary cyclist is passed by another cyclist riding with a constant speed of 3.4 m/s.
Immediately as the other bike passes the stationary bicyclist hops on his bike and accelerates at 2.4m/s2 until he catches up.
(a) How much time does it take to catch up and how far has he traveled in this time?
:
Let t = time required for him to catch up
then
3.4t = distance (in meters) traveled by both bikes when this happens
:
2.4t = cyclist speed after t seconds
= av velocity of cyclist over this distance, (from 0 to 2.4t m/s^2)
:
*t = = 1.2t^2 distance traveled over t time
It is equal to the other rider's dist
1.2t^2 = 3.4t
1.2t^2 - 3.4t = 0
Divide both sides by 1.2t
1.2t(t - 2.83) = 0
t = 0, start time
and
t = 2.83 seconds to catch up
dist = 2.83 * 3.4 = 9.622 meters
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