Question

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

Answer 1

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