Question

I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees biker B 34m ahead on the road travelling 4.7m/s. How far will biker B get before biker A catches him?

Answer 1

D=rt
when biker A catches biker B, the time they've been riding is the same, so 
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so 
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters