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Problems with kinematics equations

  1. Sep 8, 2006 #1
    I have a few homework questions involving some kinematics equations, vectors, etc. I was hoping I could some help with them.

    My first question:

    An arrow strikes a target in an archery tournament. The arrow undergoes an average acceleration of 1.37 x 10^3 m/s^2 [W] in 3.12 x 10^-2 s, then stops. Determine the velocity of the arrow when it hits the target.

    I tried this, and my answer didn't match the one in the book, I started with this:

    v_2 = v_1 + at

    And I wasn't sure this was even the right equation, or how to stick the numbers in. Time and acceleration were easy to substitute in, but then I didn't know what to do with the two velocities, one of which, I'm assuming is 0 m/s.

    Another problem I have reads as follows:

    Starting with the defining equation for constant acceleration and the equation for displacemnt in terms of average velocity, derive the constant aceleration equation

    a) from which final velocity has been eliminated
    b) from which initial velocity has been eliminated

    I got about as far as picking the equations that it says to start with:

    a = (v_2 - V_1)/t and v = d/t

    I''m not sure what to do next, please help me figure this out.

    Another one went as follows:

    A watercraft with an initial velocity of 6.4 m/s [E] undergoes an average acceleration of 2.0 m/s^2 for 2.5 s. What is the final velocity of the watercraft?

    I got confused by this, because the acceleration is perpendicular to the velocity, and so I wasn't really sure how to start, any help would appreciated.

    Lastly(I hope), is this one:

    A hockey puck rebounds from a board as shown in figure 16. The puck is in contact with the board for 2.5 ms. Determine the average acceleration of the puck over the interval.

    Figure 16 shows the puck hitting the board with a velocity of 26 m/s 22 degrees from the board, and then rebounding off at the same angle with a velocity of 21 m/s.

    I really wasn't sure how to go about starting this, so any guidance here would be appreciated.
  2. jcsd
  3. Sep 8, 2006 #2


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    Staff: Mentor

    Looking at the first problem
    one has to be careful when using formulas. One must distinguish between absolut time and duration (or difference in time).

    Let v2 = v (t2) and v1 = v (t1), then v2 = v1 + a(t2-t1), where a is the average acceleration between t2 and t1. Let 1 reference the initial state and 2 reference the final state.

    When the arrow stops, v2 = 0, so the previous equation becomes, v1 + a(t2-t1) = 0

    Then v1 = -a(t2-t1)

    Now the problem states "The arrow undergoes an average acceleration of 1.37 x 10^3 m/s^2 [W] in 3.12 x 10^-2 s, then stops."

    So, (t2-t1) = 0.0312 s, and the acceleration should be a deceleration, or acceleration of -1379 m/s2, so

    v1 =


    See if this helps.
  4. Sep 8, 2006 #3
    43 m/s [E], right? My textbook says 42.8, not sure how that came about though.
  5. Sep 8, 2006 #4


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    Staff: Mentor

    I got about 43 m/s. I see the mistake on my part, I wrote 1379, instead of 1370 m/s2. Using 1370, velocity is 42.7 m/s. My apologies for the error. :redface:

    I take [E] and [W] refer to east and west, respectively.
    Last edited: Sep 8, 2006
  6. Sep 8, 2006 #5
    Ok, thanks for the help with that question. Would you be able to help me with the others as well?
  7. Sep 9, 2006 #6


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    Staff: Mentor

    We are pleased to assist, but please show some work on each problem, e.g. write the formulas and substitute in the values.
  8. Sep 9, 2006 #7
    Subsituting is the problem, for my second question. I wasn't sure how to substitute everything in, since it's a derivation, and a little complicated, I think. I'm pretty sure my forumlas were right, I'm just not sure what to do with them.

    For my third question, I tried to separate the acceleration into x and y components, and since it's direction South, it only has a y component. This means that the acceleration caused by it is only in the South direction, so I'm guessing that also means that the velocity is change is also in the South direction. The problem is, I don't know how to show the calculate the change in velocity, I was thinking of this:

    V_2 = V_1 + a(t_2 - t_1)

    And then using zero as the initial velcoity, but that didn't make sense to me, for some reason. The acceleration and time were easy to substitute in, but I'm not sure if this is right, or how to explain it.

    As for my third question, I guess 26 m/s [22 degrees S of E] would be the initial velocity, whereas 21 m/s [22 degrees N of E] would be the final velocity, and then the change in time is given to be 2.5 ms. The velocities are in different directions though, so would I have to break them up into their x and y components in order to work with them?

    Sorry if it's not much, but this is about all I can come up with, at least I think it's a little better than before.
  9. Sep 9, 2006 #8

    Doc Al

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    Staff: Mentor

    The equation for acceleration is fine. Express the second equation in terms of initial and final velocities. (In that equation, "v" is the average velocity.)

    Once you've done that you'll have two equations in terms of V_1 and V_2. To eliminate a variable, solve for it in one equation, then substitute it into the other.
  10. Sep 9, 2006 #9

    Doc Al

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    Staff: Mentor

    You are exactly on the right track. Hints: Does the x-component of velocity change? What is the final y-component of velocity? (What's the initial y-component of velocity?)

    It's easiest if you treat the x and y components separately. Find the x and y components of the initial and final velocities. (You don't have to break them into components, if you are comfortable finding the change between initial and final velocity vectors.)
  11. Sep 9, 2006 #10
    I have a question about the arrow problem, how do I find out the direction of the velocity? The book says East, but my calculations say West, did I do something wrong?
  12. Sep 9, 2006 #11

    Doc Al

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    Staff: Mentor

    Show what you did. I would call the positive x-axis to be east, in writing your equations.
  13. Sep 9, 2006 #12
    Yeah, that's what I did, but the acceleration is west, and the arrow doesn't seem to have an initial velocity.
  14. Sep 9, 2006 #13

    Doc Al

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    Seriously, does that make any sense at all?

    In your first post you gave the proper equation. Show how you used it.
  15. Sep 9, 2006 #14
    I meant that the arrow's initial velocity is 0 m/s, not that it was non exsistant, come to think of it, that really doesn't make any sense, since the final velocity is 0 m/s, then the overall acceleration would be 0.

    Woops, now I think I know what I did wrong, the final velocity is 0 m/s. so when you rearrange the equation the accelaration becomes negative, and since it's already in the negative direction, it becomes positive, resulting in a positive velocity.
  16. Sep 9, 2006 #15
    I just figure out the derivations, I think I'm doing something wrong. I start with these, right?

    a = (v_2 - V_1)/t and d = 1/2(v_2 - v_1)(t)

    Things never seem to work out when I try to substitute and rearrange them, could someone help me please?
  17. Sep 9, 2006 #16
    When you average numbers, you add them, not subtract them.
  18. Sep 9, 2006 #17
    Oh, I forgot about that, it's a bad habit since I'm usually calculating the difference. I'll see if I can figure it out now, thanks for reminding me about that, though I feel like an idiot now, many of my problems on individual homework questions come from carelss mistakes like that.
  19. Sep 9, 2006 #18
    I think I got it right, finally:

    a = (v_2 - V_1)/t and d = 1/2(v_2 + v_1)(t)

    2d = (v_2 + v_1)(t)
    2d/t = v_2 + v_1
    2d/t - v_1 = v_2

    a = (v_2 - v_1)/t
    a = ((2d/t - v_1) - v_1)/t
    at = (2d/t - v_1) - v_ 1
    at + 2v_1 = 2d/t
    at^2 + 2v_1t = 2d
    d = v_1t + 1/2at^2

    Now I just have to do the other one, and then the rest of the questions.
  20. Sep 10, 2006 #19
    This is one my earlier questions, the only I have yet to solve, well, there's another one in my book, but I haven't looked at it yet. I was wondering if someone could help me a bit with this. I can probably calculate the resultant vector using the cosine law, but something doesn't make any sense to me. The only time given is the time during which the puck is in contact with the board. The entire trip is more than just the time when the puck is in contact with the board, so this doesn't really make any sense, someone please help me figure this out.
  21. Sep 10, 2006 #20
    Someone please help.
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