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2nd derivative

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that for a projectile d^2 (v^2)/dt^2 = 2g^2

    2. Relevant equations


    3. The attempt at a solution
    Since we are only dealing with the y component for this. 2nd derivative would be 2. But since gravity acts on it then d^2 (v^2)/dt^2 = 2g. I don't get how to get to 2g^2. Can someone help me please.
     
  2. jcsd
  3. Nov 10, 2014 #2

    gneill

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    Start with a general expression for the vertical velocity of a projectile.
     
  4. Nov 10, 2014 #3

    Doc Al

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    Step by step. What's the first derivative of v^2?
     
  5. Nov 10, 2014 #4
    First derivative would be 2v.
     
  6. Nov 10, 2014 #5
    Vertical velocity = v - at. Im still confuse on what to do with this equation. Since dv/dt = a.
     
  7. Nov 10, 2014 #6
    If Vy=v^2-gt then the first would be dv/dt = 2v-g.
     
  8. Nov 10, 2014 #7

    gneill

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    That's not complete. In that equation v is a function of time, so apply the chain rule.

    that would be ##v = v_o - g t##. Square both sides so that you have an expression for v2. Then do your derivatives. Remember that dv/dt = g.
     
  9. Nov 10, 2014 #8

    gneill

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    I don't know where you got that expression for Vy. It's not correct.
     
  10. Nov 10, 2014 #9

    PeroK

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    It's your maths that's letting you down here. I don't think you understand what is being asked. You are asked to differentiate ##v^2##. Not ##v##.

    I would first work out what ##v^2## is. That seems logical to me: you are asked to differentiate ##v^2##, with respect to time (twice). So, let's first have ##v^2## as a function of ##t##.

    ##v^2 = \dots ##

    Can you do that?
     
  11. Nov 10, 2014 #10

    Doc Al

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    As gneill already pointed out, that is not complete.

    What you found is the first derivative with respect to v. But what you need is the first derivative with respect to t. So, keep going. (Think chain rule.)
     
  12. Nov 10, 2014 #11
    If I apply chain rule. Let u=(v-gt)^1/2. Then : dv/dt = (u)^1/2
    dv/dt = du/dt × dv/du = -g/2 (v-gt)^1/2. Am I on the right track? Then my next step would be to differentiate my answer again.
     
  13. Nov 10, 2014 #12

    Doc Al

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    Why not start with what you were given?

    Let u = v^2
    du/dt = du/dv * dv/dt.

    Hint: What is dv/dt for a projectile?
     
  14. Nov 10, 2014 #13
    du/dt = 2v * -g = -2vg.
    2nd derivative:
    du/dt = -2g * - g = 2g^2.
    Thank you. Can I ask one more question. What is wrong with my first attempt using the chain rule? Can i still get the same answer if I kept going with my solution?
     
  15. Nov 10, 2014 #14

    Doc Al

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    I couldn't follow what you were trying to do.

    Where did you get this?

    Perhaps you meant to follow gneill's advice:

    You can try that again and you should get the same answer.
     
  16. Nov 10, 2014 #15
    Yes I followed his advice. I'll give it one more try using that. Thank you.
     
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