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Pytels Dynamics 12.10: parabolic path, velocity, acceleration

  1. Jun 12, 2017 #1
    1. The problem statement, all variables and given/known data
    An automobile goes down a hill that has the parabolic cross section shown. (see image attached)
    Assuming that the horizontal component of the velocity vector has a constant
    magnitude v0, determine (a) the expression for the speed of the automobile in
    terms of x; and (b) the magnitude and direction of the acceleration.

    2. Relevant equations
    y=h(1-x2/b2)

    3. The attempt at a solution
    I have attached an image

    It gives a solution in (a) v0√1+(2hx/b2)2
    and (b) 2hv02/b2

    How do I proceed in (a) to come to the given solution?
     

    Attached Files:

  2. jcsd
  3. Jun 12, 2017 #2

    kuruman

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    Have you considered mechanical energy conservation?
     
  4. Jun 12, 2017 #3
    Conservation of mechanical energy is introduced later in pages 148-149.
    This problem is at the first pages(pg. 23) of Dynamics 2nd edition under the Rectangular Coordinates section.
     
  5. Jun 12, 2017 #4

    kuruman

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    Actually, I was off the mark initially. This is not an energy conservation problem, but a simple application of calculus. You are looking for ##v = \sqrt{v_x^2+v_y^2}##. You know that ##v_x = v_0##, that ##y = h(1-x^2/b^2)## and that ##v_y = dy/dt##. So ...
     
  6. Jun 12, 2017 #5
    so for vy = dy/dt i get vy = - (h/b2)2x

    and v = sqrt(v02 + (- (h/b2)2x)2)

    I don't know how to proceed then
     
  7. Jun 12, 2017 #6

    kuruman

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    You did not apply the chain rule of differentiation carefully enough. Try again. Remember that x depends on time.
     
  8. Jun 16, 2017 #7
    Ok. What I get is different from the given soution at the back of the book.
    It is given ##v = v_0 \sqrt {1 + \left( \frac {2hx} {b^2}\right) ^2}##
    what I get is ##v = v_0 \sqrt {1 + \left( \frac {-2hx} {b^2}\right) ^2}##
    Check also my upload
     

    Attached Files:

  9. Jun 16, 2017 #8

    kuruman

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    ##\left( -\frac{2hx}{b^2} \right)^2=\left( (-1)\frac{2hx}{b^2} \right)^2=(-1)^2\left( \frac{2hx}{b^2} \right)^2=\left( \frac{2hx}{b^2} \right)^2##
     
  10. Jun 16, 2017 #9
    Any hint in finding the magnitude of acceleration?
    It should be ##\frac {2hv_0^2} {b^2}##

    I have done the ##a_y=\frac {d} {dt}\left( v_y\right) = \frac {d} {dt} \left [v_0\left(\frac {-2hx} {b^2}\right)\right]##. How do i proceed from here?

    and ##a_x## should be zero since ##v_x## is constant
     
  11. Jun 16, 2017 #10

    ehild

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    Apply Chain Rule, (see @kuruman's Post#6)
     
  12. Jun 16, 2017 #11
    It should be as shown at the attached file
     

    Attached Files:

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