A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later,
the bicyclist hops on his bike and accelerates at 2.11 m/s2 until he catches his friend. How much time does it take until he catches his friend
1 answer:
From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.
The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:
3.63 t = 1.055 t^2 ==> 1.055 t^2 - 3.63 t = 0
==> t = 0 s or t = 3.44 s
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