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1D Kinematics

  1. May 27, 2007 #1

    Runner A is initially 3.0 mi west of a flagpole and is running with a constant velocity of 8.0 mi/h due east. Runner B is initially 1.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet (in miles)?

    2. Relevant equations

    deltax = 1/2 (v+vo)t

    3. The attempt at a solution

    I took the mileage given in the problem to be my delta x's. The velocities to be my vo. Time I converted the mi/h to hrs. Solving for v. I got 8 and I got 5. I really don't know if I am doing this problem rigth or not??
  2. jcsd
  3. May 27, 2007 #2
    Well from what it looks like all you have is two equations with one unknown right? So just solve for you one unknown.
  4. May 27, 2007 #3
    You know that their times are equal. They both run for the same time.
    t = dx/8 = dy/5
    dx is the distance first one ran
    dy is the distance second one ran
    You also know another relationship between dx and dy. Their sum is...
  5. May 28, 2007 #4
    right ... runner a's time is 3/8 h and runner b's time is 1/5 h ... their sum is 0.575 but i don't understand how i get how far the runners are from the pole from these calculations
  6. May 28, 2007 #5
    You should not have two values for time here. This isnt possible, how can 2 runners meet at different times if they are running towards each other. They should only meet once so you should have one value for t.

    Lets break this down. The problem gives you a distance east of the flagpole for runner one, it gives you the same information about runner 2 and it gives both runners velocities.
    In your eqn you have your target variables time and x(final). You will have seperate equations for each runner with a common target variables ("t") and x(final). From here you just solve the system of equations to find you time.
  7. May 28, 2007 #6
    ok so i have ...

    using the equation deltax = 1/2 (v + vo)t

    runner A: 3mi = 1/2 (8mi/h)t
    t=3/4 h

    runner B: 1mi = 1/2 (5mi/h)t
    t= 5/2 h

    total time is 3.25 h

    having both time and total velocity (8mi/h + 5mi/h) and having total distance 4mi I am lost at where to go from here??
  8. May 28, 2007 #7
    Not quite,your equation says deltax=.5(v+vo)t
    so delta x is a change in x or xfinal - xinitial so runnerA:x-3mi=.5(8mi/h)t
    runner B:x-1mi=.5(5mi/h)t
    so you have two equations with two unknowns solve for t you should get the same value of t for each runner.

    Once you find t you can find the distance each runner ran and find where they meet.
  9. May 28, 2007 #8
    I don't know why you would make it so complicated.
    Speed v is equal to the distance d over time t.
    v = d/ t
    So t = d/ v
    I think you know now why their times are equal.
    So for the first guy
    t = dx/vx = dx/8
    For the second guy
    t = dy/vy = dy/5
    The times are equal so
    dx/8 = dy/5
    d is the distance they run, not the distance to the flagpole. We don't know dx or dy.
    But one is 3 mi away and the other one is 1 mile away in the opposite direction. The distance between the two people is 1 + 3 = 4
    dx + dy = 4 (they run the distance between them)
    dx = 4 - dy
    dx/8 = dy/5
    (4 - dy)/8 = dy/5
    20 - 5dy = 8dy
    You find out dy, and then the time = dy/5.

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