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Prof. Shankar's (Phy I) Constant Motion Exercise 3 Question

  1. Apr 10, 2017 #1
    Hello everyone, this is my first post, so go easy on me! Thank you to those who are able to help.

    1. The problem statement, all variables and given/known data

    The problem is taken from Professor Shankar's Fundamentals of Physics exercises (Problem 3).

    Romeo is at x = 0 m at t = 0 s when he sees Juliet at x = 6 m

    Romeo begins to run towards her at v = 5 m/s. Juliet, in turn, begins to accelerate towards him at a = −2 m/s 2

    2. Relevant equations

    When and where will they cross?

    My personal questions relate to:

    1. My approach to the question as opposed to the given solution
    2. Some of the algebra in the solution

    3. The attempt at a solution

    Working from a position function of x(t) = 1/2at^2 + vot + xo (where a = constant acceleration; vo = initial velocity; xo = initial position)

    I took known variables and formed:

    x(romeo) = vot + xo (because we know Romeo's initial velocity (5m/s) and his initial position (0 m))

    x(juliet) = 1/2at^2 + vot (because we know Juliet's acceleration (-2 m/s^2) and initial position (6m))

    Prof. Shankar equates x(romero) to simply vt, as the position would be 0 at t = 0.

    Would my way of forming x(romeo) and x(juliet) be correct?

    Romeo and Juliet should cross when x(romeo) = x(juliet)

    Giving according to my more verbose variables:

    vot(r) + xo(r) = xo(j) + 1/2at^2(j)

    To solve we would equate one side to 0 and "move across" the required variables, to get for example:

    0 = 1/2at^2(j) - vot(r) + xo(j)

    Does it matter which way the variables get "moved across"?

    Solving 0 = 1/2at^2(j) - vot(r) + xo(j) with the given variables I get a result of:

    -3.38600... (-3.0 s rounded to the nearest integer)
    0.88600... (1.0 s rounded to the nearest integer)

    Prof. Shankar's solutions state: 1 second exactly. Would my above solution be incorrect?

    Here is his reasoning:


    Would I be correct in saying that he has amended the quadratic equation to better fit the time function t = v2 - v1 / a?

    Finally! I don't quite get the algebra behind v/a (1 +- .... ) Would someone be able to help me understand how he derived this from the initial equation?

    I hope I've been clear. Thanks again to anyone that can help!

    -- Lucas
  2. jcsd
  3. Apr 10, 2017 #2


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    Staff: Mentor

    If I understand correctly what no mean, no it makes no difference on which side you move all the variables.

    Solving how?

    What do you mean when you say that he "amended the quadratic equation"?

    Starting from the quadratic equation, you have
    t &= \frac{v \pm \sqrt{v^2 - 2 a x_0}}{a} \\
    &= \frac{v \pm \sqrt{v^2 - \frac{2 a x_0 v^2}{v^2}}}{a} \\
    &= \frac{v \pm \sqrt{v^2\left( 1 - \frac{2 a x_0}{v^2} \right)}}{a} \\
    &= \frac{v \pm v \sqrt{1 - \frac{2 a x_0}{v^2}}}{a} \\
    &= \frac{v}{a} \left( 1 \pm \sqrt{1 - \frac{2 a x_0}{v^2}} \right)
  4. Apr 10, 2017 #3
    Hi DrClaude,

    I "solved" it using the quadratic equation:

    https://www.symbolab.com/solver/equation-calculator/\frac{ 5 - \sqrt{5^{2} - \left(4\right)\left(-2\right)\left(6\right)}}{2 \left(-2\right)}

    https://www.symbolab.com/solver/equation-calculator/\frac{ 5 + \sqrt{5^{2} - \left(4\right)\left(-2\right)\left(6\right)}}{2 \left(-2\right)}

    Not sure if this approach is at all correct given the difference in answer?

    In terms of the quadratic, I had come across x = -b +- sqrt(b^2 - 4ac) / 2a but thought that by cancelling out the 2 and reducing the 4 to 2, it would help mirror the time function better and perhaps help with the working out of the problem. Wanted to see if this was the right way of thinking about things.

    For the final question, I can see that v^2 / v^2 is added (which is essentially 1) ...but I lose it at that point! I'll try and think a little more on the logic, but could you perhaps help me understand why this is done in the first place? What's the motivation? Why not stick with the previous equation?

    Thank you,

    -- Lucas
  5. Apr 10, 2017 #4


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    Staff: Mentor

  6. Apr 10, 2017 #5


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    Staff: Mentor

    I don't understand what you mean.

    This trick of multiplying by "1" is often used. Better get used to it. And I gave you all the steps from that point on.

    It's the same equation. Where to stop modifying an equation is often up to one's taste.
  7. Apr 10, 2017 #6
    Hi DrClaude,

    There was a problem with the image I used - I have updated the text so it should make sense now.

    Yes, you're right in that they are equivalent, but I wanted to know why the latter suited the problem better than the first. I suppose in terms of problem solving I wanted to understand that step as I didn't think of this and wanted to understand things a little better.

    When I use the following: https://www.symbolab.com/solver/equ...ft(-2\right)\left(6\right)}}{\left(-2\right)} I get the answer Prof. Shankar also gets. It is slightly off to https://www.symbolab.com/solver/equ...ft(6\right)}}{\left(2\right) \left(-2\right)} which leads me to believe that as you said before the method was correct? From what I can see, the a, b, c (a, vot(r), xot(j)) are correct?

    Thanks again,

    -- Lucas
  8. Apr 10, 2017 #7


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    Staff: Mentor

    ##a \neq a## :wink:

    Meaning that the coefficient ##a## is not the same as the acceleration ##a##.
  9. Apr 10, 2017 #8
    Hi DrClaude,

    Oh, yes! I did mix this up. 1/2(-2)t^2... so the coefficient should be 1/2(-2).


    I think this is correct now - so the method looked to be correct. Thank you for your help here.

    Might you be able help me see why Prof. Shankar used the second equation for the solution of the problem instead of the simple quadratic? There must be a reason for this beyond his preference for tasty equations?
  10. Apr 10, 2017 #9


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    Staff: Mentor

    It makes it more obvious which branch of the square root (+ or -) to keep, even before calculating the numerical values.
  11. Apr 10, 2017 #10
    Right! Thanks again.
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