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Vectors Question

  1. Jan 11, 2008 #1
    1. The problem statement, all variables and given/known data

    While driving on the interstate 83, you notice a friend half a mile in front of you. Assuming your friend is driving at the legal 65 miles per hour, determine how fast you would have to travel (with respect to land) to catch him in 5 minutes.

    2. Relevant equations

    [tex]V_{1}[/tex]= [tex]V_{0}[/tex] + at

    [tex]d_{1}[/tex] = [tex]d_{0}[/tex] + [tex]\frac{1}{2}[/tex]t ([tex]V_{1}[/tex]+[tex]V_{0}[/tex])

    3. The attempt at a solution


    I first set up a table to compare the final and inital velocities, the acceleration, time, and the final and initial distances.

    Then...

    *to figure out the Final distance of car 2 I did...


    [tex]d_{1}[/tex] = [tex]d_{0}[/tex] + [tex]\frac{1}{2}[/tex]t ([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
    [tex]d_{1}[/tex] = [tex]\frac{1}{2}[/tex]([tex]\frac{5}{60}[/tex])(65+65)
    [tex]d_{1}[/tex]= 5.7 miles for Car 2

    *to figure out the initial distance of Car 1...

    5.417 +1/2 = 5.917


    * to find the final velocity of Car 1...

    [tex]d_{1}[/tex] = [tex]d_{0}[/tex] + [tex]\frac{1}{2}[/tex]t ([tex]V_{1}[/tex]+[tex]V_{0}[/tex])

    0=5.917 + (1/2)(5)([tex]V_{1}[/tex]+65)
    -5.917=2.5([tex]V_{1}[/tex]+65)
    2.37=[tex]V_{1}[/tex]+65
    67.37m/[tex]s^{2}[/tex] = [tex]V_{1}[/tex]


    When my teacher returned my paper to me...all she said was I solved this problem in the method of a projectile problem...I don't know why she said this. Please help me find my mistake! Thanks so much!
     
  2. jcsd
  3. Jan 11, 2008 #2
    From the statement of the problem, I would not have thought this was meant to be an acceleration problem. I think it's just asking you to determine the constant speed at which you'd have to move in order to catch up with your friend in 5 minutes. In other words, don't consider any acceleration from an initial speed (which isn't given, after all) to this speed; just assume you're traveling at that speed when you're one half-mile behind your friend and the 5-minute clock starts.
     
  4. Jan 11, 2008 #3
    Forgive me if I am wrong.

    But I would find how far he would go in 5 Minutes at 65 Miles an hour. That's about 5.42 miles. Then just add the half mile your friend was in front of you, so 5.52 miles. Then just figure out how quickly you would need to go to go 5.52 miles in 5 minutes.

    D=VT
    5.52=x(.083)
    Thats about 66.5 MPH. That may be wrong, but it didn't seem to far out...
     
  5. Jan 11, 2008 #4
    ohspyro - The rules for this forum state that complete solutions should not be given for any problems posted here. I know you were just offering a suggestion for a solution, but the original poster could simply take your solution and use it as an answer - whether right or wrong. And by the way - it's wrong. Just a little error, but I'll let aquamarine work through it to find the mistake.
     
    Last edited: Jan 11, 2008
  6. Jan 11, 2008 #5
    alright, thanks
     
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