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Acceleration Problem.

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A bike accelerates at 1m/s^2 from rest. (0 meters per second) After 10 seconds a car accelerates at 2m/s^2 from rest.

    2. Relevant equations

    When are they going to meet, and where?

    3. The attempt at a solution
  2. jcsd
  3. Aug 23, 2010 #2
    Another piece of info.

    When the car starts accelerating, the bike is 50 meters away and has a velocity of 10 meters/sec.
  4. Aug 23, 2010 #3


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    Just FYI, it's against forum rules to post a problem without showing what attempt you've made so far. Other people doing your homework for you does not help you in any way whatsoever (in addition to being tantamount to cheating).

    Did the bike and car start from the same location? That seems like necessary information for this problem. Under the assumption that they did, how can you figure out the distance travelled by each vehicle at any moment in time? There is a simple way to do this for this type of motion under constant acceleration. Assuming you calculate that (i.e. know the positions of both vehicles at any moment in time), what fact is true about those positions when the vehicles "meet?"
  5. Aug 23, 2010 #4
    Oh, i didn't know about that sorry.

    Well they indeed start from the same location.
    I have been trying to solve this problem for the last 25 minutes but I can't find a soultion. I have been trying with these formulas:

    Final Velocity = (Acceleration)(Time)
    Distance = (Final Velocity/2) Time
    Distance =(Acceleration x Time^2)/2
    Final Velocity^2 = 2(Acceleration)(Distance)

    (The formulas include Starting Velocity, but as it is ZERO, I just ignored them; also if you don't understand something please let me now, I'm not used to write Physics in English)
  6. Aug 23, 2010 #5
    That's maybe what I'm missing, what is that fact?
  7. Aug 23, 2010 #6


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    Right. So with this third formula, since you know the acceleration of the vehicle, you know how far it has travelled at any time. (Meaning that if you substitute a specific time value into the equation, you can calculate the distance travelled at that time).

    Not necessarily. The easiest way to solve this problem is to start at the instant the car begins to move. That's t = 0. This means that, at t = 0, the bike has some initial position (50 m) and some initial velocity (10 m/s). You can use those in the equation for the distance of the bike vs. time.

    If we are both at the starting line of a race and I start running first, and then you start running later, but you're going faster, then you are going to catch up to me. At the point when you catch up to me, our distances from the starting line are the same, right? In other words, our positions are equal. The same thing is true when car catches up to the bike. The two vehicle's positions are equal at that instant.
  8. Aug 23, 2010 #7
    Yes yes, I already knew that, that's the way I'm handling the problem; but after that I dont know what to do after that :S

    I have tried substituting those values on all formulas but it seems I'm missing something!
  9. Aug 23, 2010 #8


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    As far as I know, you have two equations and two unknowns. This means you can arrive at an exact solution. If you have indeed equated the two expressions for the distance of the car and distance of the bike, then you should be able to solve for the time value t at which the equality holds true. Once you know t, then you can compute d. Once you equate the two expressions, all of the physics is done. Short of actually doing the algebra for you, there is no real further help anyone can provide. If you are having trouble with the algebra, you could always post it to see if people can track down any errors.
  10. Aug 24, 2010 #9
    draw graphics it will be useful.
  11. Aug 24, 2010 #10
    Thank you very much man!

    This will be very useful, I'll go alone from here!

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