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V^2 = vi^2 + 2ad, is v velocity or speed?

  1. Jan 15, 2011 #1
    1. The problem statement, all variables and given/known data

    A student throws a set of keys vertically upward to her sorority sister, who is in a window 2.00 m above. The second student catches the keys 2.30 s later.

    (a) What was the velocity of the keys just before they were caught?

    I did y(t) = 2 - 4.9t² + v₀t

    2 = 0 - 4.9(2.3)² + v₀(2.3)

    v₀ = 12.1m/s

    Now I have two formulas, v² = v₀² + 2ad and v = v₀ + at, but they give me different answers.

    v² = v₀² + 2ad

    v² = (12.1)² + 2(-9.8)(2)

    v² = 107.21

    v = 10.4m/s

    v = v₀ + at

    v = 12.1 - 9.8(2.3)

    v = -10.4m/s

    I understand the square root will give us a negative as well, but the question is, HOW DO WE KNOW which to take? I hadn't use the v = v₀ + at, I wouldn't know whether to take positive or negative.
  2. jcsd
  3. Jan 15, 2011 #2

    Andrew Mason

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    v= v0 + at

    to start. To find the distance, integrate this with respect to time:

    [tex]d = \int vdt = \int v_0dt + \int atdt[/tex]

    What is a? (be careful about the sign). That will leave you with a quadratic equation with two possible solutions, only one of which is real.

    Since the question asks for velocity, you have to give magnitude and direction of v.

  4. Jan 15, 2011 #3


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    If you don't want to go the calculus route, you can keep the [tex]\pm[/tex] on the answer to [tex]v[/tex]. One answer is simply going to be unphysical, you don't know right away which to use, you just have to figure out which is consistent with the other equations and the situation.
  5. Jan 15, 2011 #4
    Does that mean the v in v² = v₀² + 2ad is actually speed and not velocity?
  6. Jan 15, 2011 #5


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    It's velocity. The problem is when you square something, you lose it's directionality (whether it's positive or negative). For example, whether or not v = 2m/s or v = -2m/s is lost in that equation since [tex]v^2 = 4{{m^2} \over {s^2}}[/tex] for either case.
  7. Jan 16, 2011 #6
    I had the same thoughts. I haven't looked at Newtonian mechanics in decades. So where does the bogus solution come from?

    v² = v₀² + 2ad

    (1/2)mv² - (1/2)mv₀²= adm

    Delta K + Delta P = 0

    The change in kinetic energy is equal to the negative change in gravitational potential energy. But, then I see that both solutions are valid. We can throw a ball upward and it hits the ceiling elastically, so its velocity is sign inverted but the energy balance is still conserved. So either solution is valid--not bogus at all, but only one solution is valid for freely falling objects.

    The OP has given us a challenge: Can you determine which solution is the right one without knowledge of differential or integral calculus? This is a precalc question.
    Last edited: Jan 16, 2011
  8. Jan 16, 2011 #7


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    Newton's Laws are time-reversal symmetric. You could run time backwards and nothing would seem out of the ordinary in terms of physics (a person dropping a ball off a cliff could be seen as someone having a ball thrown up at them). So you have to go based on what you know physically what must happen.
  9. Jan 16, 2011 #8
    OK, can you use your time reversal symmetry to recover the initial conditions...

    An object is tossed vertically upward to you. Did you catch it as it was coming up at you at 10.4 m/sec, or did you catch it after it passed you and comes back down at -10.4 m/sec?

    This is the dilemma of flyingpig. You don't get to use calculus. You are just given a canned set of equations.
    Last edited: Jan 16, 2011
  10. Jan 16, 2011 #9


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    Not exactly, although I missed the point too. What you're saying is what happens when you have 2 final velocities. This happens when you aren't given the time. In this case, we know the time, but this will result in 2 different initial velocities. Actually what I said about the answers being unphysical is not correct. Unphysical means impossible but in actuality, the two results really only mean you have 1 solution that isn't in the domain of the problem (ie. a negative time, or a "return" trip if the question were about the keys being thrown up and caught as soon as possible). So yah, not unphysical, just impossible to determine which is needed simply through the math.
  11. Jan 22, 2011 #10


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    The solution to the equation v2 = 107.21 is v = ±√(107.21).

    That is, there are TWO possible answers.

    This can be shown in a number of ways.

    Two are:

    1: v2 = 107.21

    → v2 – (√(107.21))2 = 0

    → (v – √(107.21))·(v + √(107.21)) = 0

    → v = √(107.21), or v = –√(107.21)

    2: v2 = 107.21

    → √(v2) = √(107.21)

    → | v | = √(107.21)

    → v = √(107.21), or –v = √(107.21)

    Yes, it's true that [tex]\sqrt{x^2}=\left|x\right|\,.{[/tex]
  12. Jan 22, 2011 #11
    But to decide which, that is my question.
  13. Jan 29, 2011 #12
    Oh my god, I just realized something.

    The v in [tex]v^2 = v_{0}^2 + 2a\Delta_{x}[/tex] is speed.

    And I have a cheap proof

    Let F = ma and therefore F/m = a

    Then we have [tex]v^2 - v_{0}^2 = 2\frac{F}{m}\Deltax[/tex]

    [tex]\frac{1}{2}m(v^2 - v_{0}^2) = F\Delta x[/tex]

    [tex]\Delta KE = F\Delta x[/tex]

    clearly, we know that v in KE must be speed and this is why [tex]v^2 = v_{0}^2 + 2a\Delta x[/tex] will never give me the answer.

    Am I right or what? Cmon!
  14. Jan 29, 2011 #13


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    No. The v in [tex]{{mv^2}/2}[/tex] is the velocity as well. V is velocity, that's it. Stop trying to convince yourself that it is speed. Use the other kinematics you know.

    [tex]v(t) = v_0 + at[/tex]

    The acceleration can be positive or negative. Speeds are strictly positive. If v(t) were indeed a speed, how could you ever account for what will eventually happen with all negative accelerations?
  15. Jan 29, 2011 #14
    But that thing [tex]{{mv^2}/2}[/tex] is Kinetic energy, v is speed, it says so in my book
  16. Jan 29, 2011 #15


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    hi flyingpig! :smile:
    for a given vo, both solutions are correct …

    if the lower student throws the keys upward with speed vo, the upper student can catch them on the way up or on the way down …

    the speed v will be the same, won't it?! :smile:

    if you only use v2 = vo2 + 2ad, you have no input as to t, so both solutions are correct

    to eliminate one, you need to specify t, which means using v = vo + at :wink:
  17. Feb 15, 2011 #16
    Sorry for bringing it up after so long.

    the [tex]\frac{mv^2}{2}[/tex] is KE, it must be speed, velocity is a vector and multiplying two vectors must mean you must take either a dot or cross product, and I don't think KE involves either, so it must be speed.

    A breakthrough!
  18. Feb 15, 2011 #17
    But my book says it is wrong...
  19. Feb 15, 2011 #18


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    I'm going to pop a vain. It IS velocity, but if we call speed S, [tex] S = |{\bf V}|[/tex]. [tex] S^2 = |{\bf V}|^2[/tex]. However, since the velocity V is real, [tex] S^2 = |{\bf V}|^2 = V^2[/tex]. Same thing. It's velocity.
  20. Feb 15, 2011 #19
  21. Feb 16, 2011 #20


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    hi flyingpig! :smile:
    yes, it's wrong

    i said that for a given v0 (without specifying t), both solutions are correct …

    you need to specify t also, to choose the correct solution :wink:
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