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Basic kinematic problem, find velocity using certain data

  1. Apr 20, 2017 #1
    Hi. Sorry, couldn't make a more specific tittle, I'm not used to solve physics problems.
    A friend of mine handed me the following problem:

    1. The problem statement, all variables and given/known data
    You have a graph that denotes the aceleration of an object as a function of the position of the object. This object has a mass of 10 Kg.
    What's the speed when the position is equal to 4 meters?

    2. Relevant equations
    $$ r(t) = a \dfrac{t^{2}}{2} + v_{0}.t $$
    $$ F=m.a $$


    3. The attempt at a solution
    What I consider I did right:
    I have translated the problem to the following one, I guess it's right but it's too complicated:

    You have a object that's initially still, and then you accelerate it at 6m/s^2 for certain time t, until it goes 2 meters. Then, you the acceleration is linear and I wrote it as 6m/s^2 times (4m-r(t))/2m times 6m/s^2 where r(t) denotes the position as function of the time.
    First step to solve it, as the acceleration is constant in the beggining:
    $$ 6 \frac{m}{s^{2}}.\dfrac{t^{2}}{2} = 2m $$
    $$ t^{2} = \frac{2}{3} s^{2} $$
    $$ t = \sqrt{ \frac{2}{3} } s = \dfrac{\sqrt{6}}{3}$$
    Also using this we know that the speed of the object when it's at 2 meters of it's original location
    $$ \sqrt{ \frac{2}{3} } s 6 \frac{m}{s^{2}}=2. \sqrt{6} m/s $$

    Where I think I got things messy:

    So I make the acceleration as function of time, we have
    $$a(t) = \frac{4m-r(t)}{2m} \cdot 6 \frac{m}{s^{2}} $$
    $$a(t) = 12\frac{m}{s^{2}} 3-r(t) \frac{1}{s^{2}} $$
    Drop the units and remplace a(t) for r''(t)
    $$r''(t)+3r(t) = 12$$
    $$r( \dfrac{\sqrt{6}}{3} ) =2 \qquad r'(\dfrac{\sqrt{6}}{3})=\dfrac{\sqrt{6}}{3}.6=2 \sqrt{6} $$
    Solved the initial value problem, the solution which I think is valid when r(t) is between 2 and 4 is:
    $$ 2 \cdot \left( \sqrt{2} . \sin \left( \sqrt{2} - \sqrt{3} \cdot t \right) +cos( \sqrt{2} -\sqrt{3} \cdot t ) -2 ) \right) $$
    Put r(t)=4 and got
    $$t_{4} = \dfrac{ \left( arctan \left( \dfrac{ \sqrt{2} }{2} \right) + \sqrt{2} \right) \sqrt{3} }{3} $$

    So r'(t_{4}) is 6m/s, but I'm sure there's a better way to solve it, but I have too little physics training to find out by myself.

    Any ideas?
     

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    Last edited: Apr 20, 2017
  2. jcsd
  3. Apr 20, 2017 #2

    gneill

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    Staff: Mentor

    I think you got there in the end (I didn't check your math very closely), but there is indeed an easier way to get there.

    Consider an energy approach. On a graph of force versus distance the area under the curve represents the work done. You should be able to practically read off by inspection the total work done on the mass.
     
  4. Apr 20, 2017 #3

    haruspex

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    The physicist's approach would be to consider energy. How do you get the KE from the force applied as a function of distance covered?

    You do not show how you solved the differential equation. Did you not get an expression for ##\dot r^2## at some point? From there you can go straight to the answer without finding r(t). This is equivalent to the energy approach.

    Edit: pipped by gneill.
     
  5. Apr 20, 2017 #4
    Thank you! It's supposed to be homework of a first course of physics, and they didn't had differential equations so I was sure there was a way to work around it.
    So the total work is
    $$ 2m \cdot 6 \dfrac{m}{s^{2}} +\dfrac{(2m+0m)}{2} \cdot 6 \dfrac{m}{s^{2}} = 18 \dfrac{m^{2}}{s^{2}} $$
    I'll see if I remember well. Work was the energy invested in accelerating something you accelerated a certain distance.
    So
    $$18 \dfrac{ m^{2} }{ s^{2} } = \dfrac{ v^{2} }{2} $$
    $$36 \dfrac{ m^{2} }{ s^{2} } = v^{2} $$
    $$6 \dfrac{m}{s} = v $$
    Thank you very much!
     
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