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Kinematic Equations, solving a quadratic gives two solutions?

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Hello, after searching the internet I found this very active community and hope you can help me with my problem.

    On the Khan Academy website (Which I highly recommend) there are exercises such as Kinematic Equations. I was asked to solve Vi (Initial velocity) so I used the following equation (Screenshot also includes the question and how the answer is worked out by Khan):

    See Screenshot https://picasaweb.google.com/lh/photo/nwDuyEjZTsXZd6JbDi7V-tb5egBDP4LQWFyAOijQ15s?feat=directlink" [Broken]

    This could have come out as both negative or positive as the answer was a square root, my question is how do I determine which answer is right? With time its obvious that the answer must be positive but with acceleration or Initial velocity I do not see a way (As shown in the screenshot where I entered the answer with the wrong sign).


    2. Relevant equations

    Relevant equation is demonstrated in the https://picasaweb.google.com/lh/photo/nwDuyEjZTsXZd6JbDi7V-tb5egBDP4LQWFyAOijQ15s?feat=directlink" [Broken].

    3. The attempt at a solution

    The https://picasaweb.google.com/lh/photo/nwDuyEjZTsXZd6JbDi7V-tb5egBDP4LQWFyAOijQ15s?feat=directlink" [Broken] above shows both the question and my attempt at a solution.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 26, 2011 #2

    gneill

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    Staff: Mentor

    Hi Cones, welcome to PF.

    When faced with a choice between solutions that pop out of the mathematics, the first thing to do, as you've noted, is to discard physically impossible solutions like negative or imaginary times (although sometimes times can be negative if you're "postdicting" an event!).

    After that you're faced with choosing the solution fits the physical problem as stated if there are constraints that would eliminate one or the other of the choices. If there are no such constraints, then it is entirely possible that BOTH solutions are allowed.

    For example, in a simple projectile problem where an object is launched from the origin and lands some distance away, there are two equally valid pairs of launch angle plus launch speed that will achieve the goal; the object follows a different trajectory in each case, but still lands on the given target spot.
     
    Last edited by a moderator: May 5, 2017
  4. Jul 27, 2011 #3
    First of all thanks for the very fast response.
    I have come to the same conclusions that you've stated but I am unsure of how to apply that to the exercise, the question only accepts one answer (as shown in the screenshot) but cannot tell in this particular instance how I should have determined the right answer. Acceleration(a) or distance(d) is easy to determine when given initial velocity(Vi) and final velocity(Vf) via this process based on the change in velocity (such as whether it is negative or positive) but I don't see a method that allows me to check the Vi's possible answers against the variables given in the question.

    I hope I explained that right, once again thanks for the help.
     
  5. Jul 27, 2011 #4

    gneill

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    Staff: Mentor

    I understand the situation as you've described it. All I can suggest is that, for these contrived examples that aren't tied to a real physical situation where you'd know more about the allowed or at least the expected motions, choose the simplest result. In this case assume unidirectional constant acceleration motion -- that is, choose the positive root.

    Any ambiguity in a problem is the fault of the problem's author, not the problem solver!
     
  6. Jul 27, 2011 #5

    PeterO

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    Homework Helper

    The two answers - positive and negative usually come about when you fire a projectile from a position higher than where it lands - such as off the top of a building, but landing on the ground.

    Look at it like this:

    Suppose a projectile was fired up at an angle from the top of a tall building at time t=0, and hits the ground some time later.

    If you were watching this from a [safe] distance, side on, you could observe this easily.

    Suppose instead, two identical projectiles were fired at the same time - you would see them travel along identical paths, though would expect that one of them was slightly closer to you than the other.

    Suppose instead two projectiles were fired, one from the roof, and the other from ground level behind the building.
    Due to the skill and organisation of the two launch crews, you still saw the two projectiles traveling along identical paths much the same as in the example above.
    The projectile launched from the ground will have been fired at a different angle, and with a higher speed to achieve this, and will have been fired earlier, so as to pass the top of the building just as the rooftop launch took place.

    Since you decided to define t=0 at the time the rooftop launch took place, the other launch which took place a little earlier - at a negative time.

    Your use of equations only calculates when the projectile is at ground level - the projectile from behind the building explains the "negative answer".

    I hope you were able to read all the way through to here - and understood what I was trying to say.

    Peter
     
    Last edited by a moderator: May 5, 2017
  7. Jul 28, 2011 #6
    Thanks I managed to complete the exercise and this all helped! I'm very tempted to watch more Physics videos on that site once I've finished the other exercises now.
     
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