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Constant acceleration and Free-Fall

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data

    An object traveling along the x axis at constant acceleration has a velocity of +10 ,m/s when it is at x = 6.0 m and of +15m/s when it is at x = 10m What is it acceleration?

    B)An object projected vertically upward with initial speed of v0 attains a maximum height of h above its launch point. Another object projected up with initial speed of 2v0 from the same height will attain a maximum height of a) 4h, b) 3h, C) 2h, d) h (air resistance is negligible)

    2. Relevant equations

    3. The attempt at a solution

    A) Does this question ask for the average acceleration or instantaneous?
    Can I apply the avg. acceleration formula with this problem?

    B)For question B, Can someone explain what the problem is asking me to solve, Im confuse at the question.

    Thanks for your help.
  2. jcsd
  3. Feb 3, 2008 #2
    You might have problem finding an average acc since you got no time. You should probably refer to

    v^2 = v(initial)^2 + 2*a*(final distance - starting distance).

    Solve for a and that should be yur answer.

    For problem b you want to find how much higher object 2 can go if it's shot twice as fast as object 1.

    In other words, how much higher will object 2 go than object 1.
  4. Feb 3, 2008 #3
    v^2 = v(initial)^2 + 2*a*(final distance - starting distance).
    What equation is this and how you get it?

    And for problem B, Im still confuse on how would I approach this problem. Can someone hint me on what should be the starting point be.

    Thanks again.
  5. Feb 3, 2008 #4
    When the acceleration is constant (as in this problem), the average acceleration and the instantaneous acceleration are the same. So you can use the average acceleration formula, but you would need to find the time.

    For the second problem. it is telling you that something is launched at a certain velocity (Vo) and it reaches a certain height, h.

    It wants to know: if you launch something twice as fast (2*Vo), how much higher will it go in terms of the first objects height. Twice as high (2h), and so on.

  6. Feb 3, 2008 #5
    You got x-x0 = v0t + 1/2at^2
    and v= v0 + at.

    I think you solve for t in the bottom equation and then replace t with the new equation in the top equation. If that makes sense...
  7. Feb 3, 2008 #6
    How would I find the time? the problem only shows velocity and position of x.

    Problem B) Wouldn't it be 2h if is twice the velocity, it would reach twice the height.
  8. Feb 3, 2008 #7

    Im sorry, I don't, can you elaborate alittle more please. Thanks.
  9. Feb 3, 2008 #8
    Well, I notice that you didn't fill in the 'Relevant Equations' portion of the form. You need to take a look at the equations that you know, and relate them to the problem. You will probably end up with more than one equation and more than one unknown, which you will have to solve.

    For B, that's a nice guess, but probably not right. Here's a hint to get you started: Velocity will be zero when the object reaches its maximum height. So find an equation that relates velocity and time. Ask yourself how much more time it takes to get to the the maximum height for the second object.

    Then find an equation that relates time and distance. Use that to solve the problem.

  10. Feb 3, 2008 #9
    For both of these problems you can use the equation:

    [tex]{v_f}^2 = {v_0}^2 + 2as[/tex]

    Where v = final velocity
    [tex]v_0[/tex] = initial velocity
    s = change of distance

    This is one of the standard constant acceleration formulas. But it has a caveat - you cannot use it if the direction changes (which it doesn't for either of these questions).

    For B, don't assume just because the velocity is double that the height will double. Solve the equation as is for h. Then use 2v and solve it again for h. You will see a difference which will give you your answer.

    Edit - sorry Sheldon, you must've posted while I was writing this, and I'm not making him work as hard for the solution.
  11. Feb 3, 2008 #10
    As for your question on how to derive the equation I gave you. You can do this if you every forget. Takes a minuet or two but it beats memorizing it.

    So we have v = v0 + at and x-xo = v0t +1/2 * a * t^2

    Take the first equation and do this

    (v - v0)/a = t

    Then go to the second equation and do this.

    (x-x0) = v0(v-v0) + 1/2 * a * (v - v0)^2.

    Expand everything else and start reducing things. When you finish you get v^2 - v0^2 = 2a(x - x0).
  12. Feb 4, 2008 #11
    If first the object is at x = 6.0m , and velocity is +10m/s and later is at x = 10.0m and velocity +15m/s, and the acceleration is constant then is the acceleration 1.25?
  13. Feb 4, 2008 #12
    Nope. You're off by an order of magnitude and change.
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