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When will bike with constant velocity overtake car starting ahead with constant acceleration?

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A motorbike is driving at a constant velocity of 25m/s, and 80 meters in front of it, a car is accelerating from standing still, with a constant acceleration of 3m/s2. When will the bike overtake the car?

    2. Relevant equations
    s=vt
    Possibly s=ut+0.5at2

    3. The attempt at a solution
    I've found that the bike reaches the car's starting point after 3.2s, at which point the car is now 15.36m ahead and has a velocity of 9.6m/s. I don't know how to proceed from here other than to keep calculating how long it takes for the bike to get to the point the car is in, and repeat until they are at the same spot. Surely there must be a more practical way.
     
  2. jcsd
  3. Oct 8, 2014 #2

    SteamKing

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    Well, it helps to draw a sketch of the problem first, before blindly plugging and chugging numbers into some formulas. A little analysis before doing the calculations goes a long way.

    The problem statement asks you when the bike will overtake the accelerating car. If you were actually observing this situation, how would you judge when the bike overtook the car? Would you look at your watch, or would you look at something else which would tell you instantly that the bike had caught up to the car?
     
  4. Oct 8, 2014 #3

    jedishrfu

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    The car and the bike travel the distance so start with the equation

    Car dist = bike dist + 80

    Now plug in what you know about the car dist and bike dist as a function of time and solve for time.
     
  5. Oct 8, 2014 #4
    SteamKing: I guess I should've mentioned that I've thought about distance, and that I don't know how to find where they meet either.

    jedishrfu: By plug in, do you mean into the second equation I listed? I've thought about this for a good while now, and I just can't figure out how to apply the 80m distance to the problem.
     
  6. Oct 8, 2014 #5

    jedishrfu

    Staff: Mentor

    They start at t0 and they meet at t1 so they take the same amount of time to get to the meeting place right?

    The car travels its car dist and the bike travel its bike dist but it's 80m ahead so then you get the equations:

    Car dist = bike dist + 80

    Right?
     
  7. Oct 9, 2014 #6
    Yes, I understand that the car dist is bike dist + 80 because it's starting out 80 meters ahead, but I don't understand how that can help me find where the bike overtakes the car.

    The overtake point will be at a distance driven by the bike, which will be the same as the distance driven by the car +80m. And t will be the same for both. I really feel like I'm overlooking something simple here.
     
  8. Oct 9, 2014 #7

    PeroK

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    You don't seem to have confidence in the mathematics of two simultaneous equations. You've summed up the problem and the solution, but now you have to hit it with some mathematics!
     
  9. Oct 9, 2014 #8

    NascentOxygen

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    I think I would write bike distance = car distance + 80

    where distance means the distance each must travel from its respective starting point until they meet.

    Let their common time of travel be t seconds.
     
  10. Oct 9, 2014 #9

    SteamKing

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    That's why drawing a simple sketch is often helpful. It shows how the distance traveled by the car relates to the distance traveled by the bike (the two vehicles are not starting from the same point, for example). But, like PeroK says, at some point, you've got to stop being Hamlet and start calculating. It's OK to make a mistake; you back up and try a different approach; that's how you gain experience.
     
  11. Oct 9, 2014 #10

    jedishrfu

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    Yes you're right I misread the problem. Also I was thinking of a bicycle not a motor bike.
     
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