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I think this problem is correct: acceleration upward at a frat house

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data

    A student throws a set of keys vertically upward to his fraternity brother, who is in a window a distance h above. The brother's outstretched hands catches the keys on their way up at a time t later. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (Answer should be in terms of h, t, and g.)


    2. Relevant equations

    [itex]\Delta[/itex]x = v0t + (1/2)at2
    v2 = v02 + 2a[itex]\Delta[/itex]x

    3. The attempt at a solution

    Question 1: when did the act of throwing begin? When the keys were released from the hand, or during the wind up of the arm/hand, before releasing the keys? The physics tutor in the math lab told me the initial velocity was 0. But that is not true if at the moment of release the keys were already at a higher velocity.

    Question 2: what does "[a]nswer should be in terms of h, t, and g" mean? Do they want actual numbers using h, t, and g; or is this a numberless problem?

    a) [itex]\Delta[/itex]x = v0t + (1/2)at2

    v0 = (1/t)([itex]\Delta[/itex]x - (1/2)at2)

    v0 = (1/t)(h - (1/2)gt2)

    v0 = (h/t) - ((gt)/2)


    b) v2 = v02 + 2a[itex]\Delta[/itex]x

    v2 = ((h/t) - (gt)/2)2 + 2gh

    v = +(h/t) - (gt)/2 + 2gh and v = -((h/t) - (gt)/2 + 2gh)

    Since they keys were thrown upward and caught, v in this case must be the positive answer above.

    Thanks for any help!
     
    Last edited: Aug 31, 2013
  2. jcsd
  3. Aug 31, 2013 #2

    Doc Al

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    Staff: Mentor

    You are correct. By "initial velocity" they mean the speed with which they left the hand.

    They want you to express the answer in terms of those variables. You are not given any actual numbers.

    Careful with signs. If up is positive, what's the acceleration?
     
  4. Aug 31, 2013 #3

    Uh oh, thanks. The problem has now become much harder if the keys had a velocity great than 0 when they were released.


    Acceleration is negative? I think that negative acceleration is deceleration, but I am not sure.

    I also think positive velocity means the object is moving up or to the left, and negative velocity is the opposite. But I don't know for sure.
     
  5. Aug 31, 2013 #4
    I want to think that acceleration and velocity must always have the same sign.

    Velocity is L/T and acceleration is (L/T)/T.

    T is always positive (there is no such thing as negative time) and L has to either be positive or negative, but not both at the same time. So acceleration and velocity always have to agree in sign (either both are positive or both are negative)?
     
  6. Aug 31, 2013 #5

    Doc Al

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    Why do you think that? If the keys had a speed of zero when released they wouldn't even leave the hand, much less rise up to a distance h.

    Yes, the acceleration of gravity is downward and thus negative. In terms of g, what would it be?

    I don't think the term "deceleration" is very useful.

    In this problem, take up as positive. (We only have to deal with the vertical.) So a positive velocity means moving upward.
     
  7. Aug 31, 2013 #6

    Doc Al

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    That's not so. Consider a ball thrown straight up into the air. It goes up, then comes down. So the direction (and thus sign) of the velocity must change. But did the direction (and thus sign) of the acceleration change?
     
  8. Aug 31, 2013 #7
    Because, honestly, physics freaks me out, and is my hardest course!

    So free fall acceleration or acceleration downward due to gravity is positve acceleration, because the velocity is getting higher [EDIT: in absolute value]. But if the object's velocity is getting slower, or going against gravity, it is negative acceleration. So g in this case is actually -g.

    Ok, so negative acceleration means the object is slowing down and the velocity is decreasing in absolute value.

    Thank you. I am going to try. It might take me some time to formulate the equation though.
     
  9. Aug 31, 2013 #8

    Doc Al

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    It can be a brain-buster, all right. Stick with it and you'll do fine. :wink:

    Think this way: Acceleration is a vector (just like velocity is a vector). The acceleration of something in free fall is downward, and our sign convention says that means it is negative. And the symbol "g" is just a positive constant (equal to 9.8 m/s^2), so that means that the acceleration is -g.

    No, negative acceleration just means that the acceleration is downwards. (At least in this problem.) Yes, when you throw something up, the speed decreases. But once it begins falling, its speed will increase again. In both cases, using our sign convention, the acceleration is negative.

    You already had the problem solved! Just change the sign of the acceleration.
     
  10. Aug 31, 2013 #9
    One more thing: I don't know if v (or vf) is actually 0, since the keys were caught, or if v is right before they were caught, which is really weird that the book asked that question, since this allows for infinite answers, as "just before" can be just about anything, like 0.00001 m/s or 0.1 m/s. There are infinitely smaller figures before the final velocity reaches 0.
     
  11. Aug 31, 2013 #10

    Doc Al

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    No reason to think the speed of the keys when they are caught will be zero. Think of vf as the speed of the keys after they rise a distance h.
     
  12. Aug 31, 2013 #11
    I will! Thanks for your encouragement! Physics and calculus so far are my favorite courses. Human biology is ok, but there is nothing mathematical about it, or at least the professor is dumbing down the course for us and not making us do the biochemistry part with mathematical equations.

    Ok, this is the part that confuses me. If negative acceleration is downwards, how is an object being thrown upwards negative? The direction is upwards, so it is positive? Or does negative acceleration mean the vector arrow is pointing downwards, due to the force of gravity coming from that way?

    Ok, thank you. In this case though, I only had to do some basic algebra to get the variables in the right place. It's always setting up the problem that is hard for me.
     
  13. Aug 31, 2013 #12
    Ok, that is a good way to look at it.

    But if a train in stopped after braking, the final velocity is 0. So shouldn't it be the same for the keys? Or maybe it is because the train is gradually slowing down to 0, whereas for the keys thrown upwards, it is instantaneous stopping?
     
  14. Aug 31, 2013 #13
    Ok, I am going to re-do the problem, and this brings up another problem:

    a) [itex]\Delta[/itex]x = v0t + (1/2)at2

    v0 = (1/t)([itex]\Delta[/itex]x - (1/2)at2)

    v0 = (1/t)(h - (1/2)(-g)t2)

    v0 = (h/t) + ((gt)/2)


    b) v2 = v02 + 2a[itex]\Delta[/itex]x

    v2 = ((h/t) + (gt)/2)2 + 2(-g)h

    v = +((h/t) + (gt)/2) - 2gh
    and
    v = -((h/t) + (gt)/2) - 2gh

    v = (h/t) + (gt)/2 - 2gh
    and
    v = -(h/t) - (gt)/2 - 2gh

    So which option for v do I chose above for the answer? The first one, since that was the same is part (a)?
     
  15. Aug 31, 2013 #14

    Doc Al

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    All that you are modeling is the motion of the keys while they are in free fall, with the only force acting being gravity. Of course, after they are caught the speed is zero. And before they are thrown the speed is zero. But those facts are trivial and not helpful in solving the problem.

    Just like when they give you a problem of a ball being dropped from some height and you are asked to find the speed when it hits the ground. Of course they mean just before it hits the ground, not after it smashes into the ground and comes to rest. (That would be a trivial problem to solve--the answer would always be zero!)

    In the case of the train, you are again solving a constant acceleration problem. In that case the final speed is zero. No other forces or accelerations are involved.
     
  16. Aug 31, 2013 #15

    Doc Al

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    Good.


    There's a simpler formula to use to solve this problem; one that will not require taking square roots.

    Careful! It looks like you are assuming that
    c2 = a2 + b → c = a + b
    No!

    This method (done correctly) will give you two answers. One for going up and one for going down. You want the one going up.

    But use a different formula and you'll solve it quicker! (Hint: A formula involving time.)
     
  17. Aug 31, 2013 #16
    Ok, thank you. I think I see it now! With the train, the final velocity of 0 is part of the overall velocity as the train goes from moving to stopping. But with a ball being dropped, the ball hitting the ground is just incidental, as without that object stopping the motion, the velocity would have gone on for longer.
     
  18. Aug 31, 2013 #17
    Oh, duh. I made another stupid algebra mistake. I guess I was too worried about the problem to actually think!

    Here it is, using a different formula

    b) v = v0 + at2

    v = (h/t) + (gt)/2 + (-g)t2

    v = (h/t) + (gt)/2 - gt2
     
  19. Aug 31, 2013 #18

    Doc Al

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    Exactly!
     
  20. Aug 31, 2013 #19

    Doc Al

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    Almost. I think you mean:
    v = v0 + at
     
  21. Aug 31, 2013 #20
    Oh, I didn't even realize there was a page 2 of this thread. I kept checking back and saw my post at the bottom of page 1, so I thought there were no new updates.

    Thank you for catching my mistake Doc Al. I have been so sloppy with physics.

    Here it is again, corrected:

    b) v = v0 + at

    v = (h/t) + (gt)/2 + (-g)t

    v = (h/t) + (gt)/2 - gt

    I don't know what I would do without this help forum!
     
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