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Homework Help: Relative distance problem.

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Two cars are parked beside each other, facing in opposite directions. Car 1 accelerates at 5m/s^2 up to a max. speed of 25m/s, and Car 2 accelerates at 4m/s^2 up to a max. speed of 24m/s. How much time has elapsed when the cars are 200m apart?


    2. Relevant equations
    v=d/t
    vf^2=vi^2+2ad
    vf=vi+at


    3. The attempt at a solution
    Car 2
    ------
    576=8d
    d=72m
    it takes car 2 6seconds to travel 72m

    Car1
    --------
    625=10d
    d=62.5m
    it takes car 1 5seconds to travel 62.5m

    So, the cars are 134.5m apart at this stage. They need to travel 200m-134.5m=65.5m before they are 200m apart, yes? The answer is 6.83 seconds, but i don't understand how they got that. please help?
     
  2. jcsd
  3. Dec 16, 2009 #2
    Why you posted another thread for the same problem I have no idea, anyways...

    You have two cars, they both have accelerations (atleast until they reach their maximum speed).

    Use the formula V = Vo + at to figure out the times it takes each car to reach its max speed.

    Once you have those figured out you can look at the distance that each car traveled in the given times.

    Once you've established the point in time at which BOTH cars have reached their max speeds THEN you can simply solve the remaining distance left to by using d = v/t. (They've both reached there maximum speeds at this point, that is to say, a=0)

    Hint for the last part,

    If I say I have one car traveling 15m/s to the left and another car traveling 10m/s to the right, how would you keep the rate at which the two objects are being seperated the same if ONE of the cars had a new speed of 0m/s?
     
  4. Dec 16, 2009 #3
    So are my times wrong above?
     
  5. Dec 16, 2009 #4
    Yes. Initially you don't have an acceleration of 0. So you can't simply use v = d/t.
     
  6. Dec 16, 2009 #5
    okay so
    Car 2
    24=0+4t
    t=6 is wrong to reach distance of 72m?
     
  7. Dec 16, 2009 #6
    You computed that it takes car 2, 6 seconds to reach its maximum speed that part is correct.

    Within those 6 seconds in which car 2 is approaching it's maximum speed it is also moving a distance, compute that distance.

    Do the same for car 1.
     
  8. Dec 16, 2009 #7
    car 2
    d=1/2(4)(6)^2
    d=72m
    car 1
    d=1/2(5)(5)^2
    d=62.5m
    Right so far?
     
  9. Dec 16, 2009 #8
    Looks good.

    So, 5 seconds have passed, car 1 has reached its max speed but car 2 hasn't (it still has 1 second to go). Can you figure out how far car 1 is going to travel in that 1 second while car 2 is still accelerating?
     
  10. Dec 16, 2009 #9
    Okay, now this is where i am confused. How do we know that car 1 has reached its max speed and car 2 hasn't? I don't understand the 1 second to go part?
     
  11. Dec 16, 2009 #10
    You computed 2 times. These are the times in which it takes each of the cars to reach their maximum speed. It took car 1 5 seconds and car 2 6 seconds.

    So after 5 seconds have gone by car 1 is at its max speed and car 2 isn't.

    How do you know this? Well you just told me it takes car 2 6 seconds to get to its maximum speed, its only been 5 seconds! We have still have 1 second to go!
     
  12. Dec 16, 2009 #11
    Oh I see, so then does this make sense for 1 second distance car 2?
    d=1/2(4)(1)^2
    d=2m
     
  13. Dec 16, 2009 #12
    No. the first distance you calculated was correct, you need to find the distance car 1 will travel in that extra 1 second.

    HINT: You're acceleration for car 1 is now 0.
     
  14. Dec 16, 2009 #13
    I think car 1 in 1 second travels d=vt. So 25m. Yes?
     
  15. Dec 16, 2009 #14
    d= v/t
     
  16. Dec 16, 2009 #15
    yes so car 1 goes 62.5 + 25m in 6 seconds and car 2 goes 72m in 6 seconds, both cars are at constant speed the rest should be easy.
     
  17. Dec 16, 2009 #16
    I have the two cars 159.5m apart. So 200m - 159.5m = 40.5m left until they are 200m apart.
    I still don't understand how they got 6.83 seconds?
     
  18. Dec 16, 2009 #17
    So in 6 seconds the cars are 159.5m apart that means they still have 40.5m left.

    Well what do we know about the cars up to now?

    Well they are both traveling at their maximum speeds (a=0) and they are going in opposite directions. Think you can use this to figure out how long the remaining 40.5m is going to take them?

     
  19. Dec 16, 2009 #18

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, you're exactly right so far. All you need to do is figure out how much longer it takes (after 6 seconds) for the separation to increase by 40.5 m. In this time, t, car 1 will travel a distance d1 = vmax1t and car 2 will travel a distance d2 = vmax2t

    The separation is the sum of these distances:

    40.5 m = d1 + d2 = (vmax1 + vmax2)t

    Solve for t and you will find out how much longer they have to go for. You already know the answer, so if you don't get t = 0.83 s, you know something went wrong.
     
  20. Dec 16, 2009 #19
    since in opp. directions then they each travel 20.25m
    t=20.25/25=0.81 and t=20.25/24=0.84
    so 6seconds plus 0.84 is 6.84seconds?
     
  21. Dec 17, 2009 #20
    They don't each travel equal distances... If one car is moving faster than the other how could they possibly travel equal distances in the same ammount of time.

    I wanted to you recognize, using your intuition, that when you have two cars moving away from each other at a constant speed it would be the same as having a car move away from a stationary point(or car, in this case) at the rate of the 2 velocitys combined!

    Forget the answer right now, sit down and think about it and see if you can understand it!
     
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