this has ruin bycyles for me plz help me?

Question

A bicyclist is finishing his repair of a flat tire when a friend rides by at 3.4 m/s. Two seconds later, the bicyclist hops on his bike and accelerates at 2.6 m/s2 until he catches his friend.

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

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

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

Answer 1

I'm assuming that his friend maintains a steady 3.4 m/s.
Then

S1=S2
S1=Vt1
S2= (1/2) at2^2
we have
Vt1=(1/2) at2^2
since t1=t2+2s
V(t2+2 )=(1/2) at2^2 let t2=t
2Vt+4V=at^2

at^2 -2Vt-4V=0
now we have to solve for t

t= (2V + sqrt((2V)^2 + 4 a 4V))/2a
t= (2V + sqrt(4V^2+ 16 a V))/2a
t= (2 x 3.4 + sqrt(4 (3.4 )^2+16 (2.6) 3.4))/(2 x 2.6)
t=(6.8 +sqrt(46.24+141.44))/5.2=3.9 sec

b) S=Vt= 3.4 (3.9 + 2)=20.0 m

c) V=at= 2.6 (3.9)=10.14 m/s