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Catch-up problem, can't seem to get it

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a herd of sheep moving at 7 m/s. My dog is at rest 5m in front of the sheep and then accelerates at a rate of 3 m/s^2. how long does it take in seconds for the bulls to catch up to the cowgirl?

    2. Relevant equations

    I have a feeling that t=square root of 2s/a is relevant. But I also think that they have to be equal to each other since they meet at some point. So the V_s is 7m/s and the S_d is 5m+S_s. I know the dog's accerlation is 3m/s^2 but I don't have time here.

    3. The attempt at a solution

    I thought about calculating the dog's time as t=square root of 2s/a with s being 5. that gave me 1.8 seconds which i assume means 1.8 seconds ahead of the sheep but the sheep are going 7 m/s so that means that in 1.29 seconds they will catch up with the dog? Thanks for any clarification from the prospective helpers.
  2. jcsd
  3. Feb 1, 2009 #2

    Doc Al

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    Staff: Mentor

    I assume that the problem is how long does it take for the sheep to catch up to the dog?

    Measure position from the sheep's starting point. What's the sheep's position as a function of time? What's the dog's position as a function of time? Set those equal to solve for when they meet.
  4. Feb 1, 2009 #3
    Ha, yes I wanted to replace the agents in question.

    I don't have a position to start with the bulls, they are just moving, and the cowgirl is just whatever the bulls position is +5m. Should i be looking for position first and then time? Im at quite a loss.
  5. Feb 1, 2009 #4

    Doc Al

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    Staff: Mentor

    Assume the sheep/bulls start at x = 0. What's their position as a function of time? It's constant speed motion.

    Given the above, the dog/cowgirl starts at x = 5. What's her position as a function of time? It's accelerated motion.
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