Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.5 m/s^2 until he catches his friend.

(a) How much time does it take until he catches his friend (after his friend passes him)?

(b) How far has he traveled in this time?

(c) What is his speed when he catches up?

Answer 1

a)

Distance traveled by friend is two seconds is

d=2*3.1 =6.2 m

Distance traveled by Bicyclist = Distance traveled by friend

(1/2)at² =6.2+Vt

(1/2)(2.5)t²=6.2+3.1t

1.25t²-3.1t-6.2=0

t=3.789 s =3.79 s

Total time

T=3.79+2 =5.79 s

b)

Distance traveled by bicyclist is

D=(1/2)(2.5)(3.79)² =17.95 m

c)

Speed of bicyclist when he catches up is

V=at =2.5*3.789 =9.4725 m/s