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Finding the distance at which Car 1 overtakes Car 2

  1. Aug 24, 2011 #1
    Problem 1:
    Two cars are traveling along a straight road. Car A maintains a constant speed of 86 km/h; car B maintains a constant speed of 106 km/h. At t = 0, car B is 30 km behind car A. How far will car A travel from t = 0 before it is overtaken by car B?


    Problem 2:
    At t = 0, a stone is dropped from a cliff above a lake; 1.5 seconds later another stone is thrown downward from the same point with an initial speed of 31 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

    These are two problems that came up on the homework, and I still am unable to solve them. I am able to get all the problems with one object involved, I just don't seem to comprehend how to set these up and solve them. Any help is appreciated.

    Equations in this chapter:
    v = v0+at
    [itex]\Delta[/itex]x=v0t+1/2at2
    vav=1/2(v0+v)


    I have tried graphing the problems to understand them, but I think I am lacking the fundamental understanding of what each should look like. I have reread the chapter and I am still in the same boat. Text book: Physics for Scientists and Engineers 5th ed.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 24, 2011 #2

    Hootenanny

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    Welcome to Physics Forums.

    HINT: When the two cars/stones overtake/hit the water, their relative displacements will be zero. In order words you are looking for the distance when their two positions are the same.
     
  4. Aug 24, 2011 #3
    Thanks for the hint and the welcome!

    For the first problem, I am assuming the acceleration is 0. Therefore, I can use the 2nd formula as delta x = (v0)t

    I believe delta x of car A would be 86km/hr (t) and delta x of car B would be 106 km/hr (t). To find where they have the same displacement, I am guessing I set it up as 86km/hr (t) = 106km/h (t) + 30km
    Solve for t, t = 30 km / (86km/hr - 106 km/hr).

    Seem to be the right track?
     
  5. Aug 24, 2011 #4

    Hootenanny

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    Yes. Spot on. Once you have the time, you then only need to work out how far car A has travelled in time t.
     
  6. Aug 24, 2011 #5

    PeterO

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    Not quite: car A is the one with the 30km head start. Your "equation" has the advantage with Car B
     
  7. Aug 24, 2011 #6

    Hootenanny

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    Good catch! I didn't spot the sign error.
     
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