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?

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