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A few Calc questions.

  1. Jun 5, 2003 #1
    On Moon:
    It would take the same ammount of time to get down as it took to go up(maximum displacement, when v = 0).

    Thus, V(t) = ds/dt = 832 - 5.2t = 0. t = 160s. It would take about 2 minutes and 40 seconds.

    On Earth:
    Same notion. V(t) = ds/dt = 832 - 32t = 0. t = 26s.

    -- Am I right?

    Because V(t) = ds/dt = 3t^2 - 12t + 9,
    a(t) = dv/dt = 6t - 12

    Particle has v = 0 at t = 3, and 1 sec

    Thus, the particle has acelleration at v = 0 at:

    a(3) = 6 m/s^2
    and
    a(1) = -6 m/s^2

    -- Am I right?


    Please correct my mistake. Thanks.
     
    Last edited: Jun 6, 2003
  2. jcsd
  3. Jun 5, 2003 #2
    #2 looks right.

    #1, a few discrepancies...

    First of all, if the position functions are really as you typed them:
    s = 832 - 2.6t^2
    and
    s = 832 - 16t^2

    then your derivatives are wrong because there is no t in the first term of each equation.

    On the other hand, the problem, as you posted it, doesn't say that s is the MAXIMUM height, and it also isn't clear whether there are two different s's, or only one value of s...

    On the other hand, there really should be a t in the first term, since the usual position function is s = v0t - .5at^2
    So should it say s = 832t - 16t^2???

    On the other hand, maybe the trick is that the t is "hidden" such that 832 = v0t and both v0 and t are unknown...

    On the other hand, that acceleration for the moon doesn't really make sense either; free fall acceleration near the surface of the moon is approx. 1.6 m/s^2...

    Too many hands.

    Bailing out...
     
  4. Jun 6, 2003 #3
    are you a teacher prudensoptimus?
     
  5. Jun 6, 2003 #4
    No, I'm a freshman in highschool. I'm gonna be in 10th grade in beginning of August, so I want to prepare myself for the class.
     
  6. Jun 6, 2003 #5
    I made a typo. Yes there are t's after 832s...





    I think it makes sense. It takes longer for things to drop on the moon because the acceleration comparing to the earth's gravity is smaller.
     
    Last edited: Jun 6, 2003
  7. Jun 6, 2003 #6
    OK, then your answer is correct.

    (and I just realized that the equations for question 1 are in feet/sec, not meters/sec, so the 2.6t^2 for the moon is OK too)
     
  8. Jun 14, 2003 #7
    You know, you can make your work really easier (no need for calculus !).
    s = 832t - 2.6t^2
    s is displacement.
    now, displacement will be equal to zero in only two cases (in projectiles) :
    1-the object didn't move yet
    2-the object went up and back down to the earth's surface.
    So all you have to do is to substitue the value of s with 0, and solve the equation :smile:.

    And BTW, if i understood ur first way of solving this right, then i think there is something wrong about it. You see the velocity of the bullet will not be 0 when it reaches the surface, it will actually be equal in magnitude (and opposite in direction) of the initial velocity.
     
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