Question

The leader of a bicycle race is traveling with a constant velocity of +12.60 m/s and is 11.4 m ahead of the second-place cyclist. the second-place cyclist has a velocity of +9.20 m/s and an acceleration of +1.50 m/s2. how much time elapses before he catches the leader?

Answer 1

Let the time be t.

so,In time t , distance travelled by 1st cyclist = 12.6 t

distance travelled by 2nd cyclist = 9.2t + 0.5 (1.5) t^2

Now, cyclist 1st is already 11.4 m ahead of 2nd cyclist.

so, 9.2t + 0.5 (1.5) t^2  = 11.4 + 12.6t

find t :

t =  6.77 sec