Ask Question
30 December, 20:45

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?

+5
Answers (1)
  1. 30 December, 22:59
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers