Physics displacement,velocity and acceleration question?

Question

The question is as follows, ;-)

Two cyclists,A and B, start a race.A accelerates for the first 5 seconds, until his velocity reaches 12 m/s, after which he travels with constant velocity. B accelerates for the first 10 seconds, until his velocity reaches 15 m/s, after which he travels with constant velocity.

a)Calculate the distance travelled by both cyclists in the first 10s.

b)Who is in the lead after 10 seconds?

Thanks in advance ;-)

Answer 1

v(average) = (v(0) + v(f))/2

for A

v(av) = (0+12)/2=6m\s

x=v(av)*t

x=6*5=30m

a=(v(t)-v(0))\t

a=12\5=2.4m\s^2

after 10 seconds

x=(1\2)*a*t^2

=(1\2)*2.4*100=120m

for B v(av) = (0+15)/2=7.5m\s

x=v(av)*t

x=7.5*10=75m

So A is in the lead after 10 seconds