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A cart driven by a propeller with a constant acceleration

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x = 0m, with an initial velocity of +4.85m/s and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of x = +11.8m, where it begins to travel in the negative direction. Calculate the acceleration of the cart.

    2. Relevant equations x=x(i)+1/2(v+v(i))Δ(t)


    3. The attempt at a solution i listed all my components and plugged into the formula then multiplied that t by the m/s to get acceleration i got 23.6m/s^2 but that is wrong
     
  2. jcsd
  3. Sep 10, 2012 #2
    Think about what is happening at the position [itex]x = 11.8\text{m}[/itex]. Is the cart accelerating? What is it's velocity?
     
  4. Sep 10, 2012 #3
    At x= 11.8m the velocity is 0 because it stops then turns the other direction. When the cart is going in the positive direction its at a constant acceleration i thought that would be zero because its not speeding up or slowing down but when it reaches x final and stops it has to have an acceleration to speed back up, right??? i am no good at this so correct me....
     
  5. Sep 10, 2012 #4
    i thought at first that the velocity would be the same on the way back just negative but when i use that number it doesnt make sense in the formula...
     
  6. Sep 10, 2012 #5
    Try using this equation: [itex]x=x_0 + v_0 t + \frac{1}{2}at^2[\itex]

    You are correct that the velocity is 0 because we KNOW that the cart has to go the other way. But what is the acceleration at [itex]x = 11.8\text{m}[\itex]? Acceleration, like velocity, can be negative. A negative acceleration is colloquially called "deceleration".
     
    Last edited: Sep 10, 2012
  7. Sep 10, 2012 #6

    Simon Bridge

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    I usually get students who are confused about this to draw the v-t graph for the motion.
    displacement is the area under the graph and acceleration is the slope of the graph ... it helps them get their heads around the motion. This one will just be a triangle.

    To use the kinematic equation method, you start by listing what you know:

    d = 11.8m
    u = 4.85m/s
    v = 0m/s
    a = ?

    Now find the kinematic equation with each of these letters in it, and none of the others.
    Note: the equation in post #5 does not do this.
     
    Last edited: Sep 10, 2012
  8. Sep 11, 2012 #7

    CWatters

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  9. Sep 11, 2012 #8
    hi. i dont understand you equation in post 5.
    i tried using the formula v^2=vnot^2+2a(x-xnot) and got 0.997m/s^2 and its still wrong
    seems like that is the only equation i can use since i do not have time...
    thanks
     
  10. Sep 11, 2012 #9
    #5

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  11. Sep 11, 2012 #10

    CWatters

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    I wouldn't use it either.

    That looks like this equation (one of the standard equations of motion)...

    V2=U2+2as.

    where

    V= final velocity = 0
    U= initial velocity
    S= displacement

    You appear to have made a trivial mistake when using it to calculate "a".

    Rearrange it carefully and you get..

    a = -U2/2s

    = -0.997m/s2

    eg not +0.997m/s2. A +ve number implies the cart is accelerating (eg getting faster). A -ve number implies it's decelerating.
     
  12. Sep 11, 2012 #11
    oh ok now i see. thanks!
     
  13. Sep 11, 2012 #12

    Simon Bridge

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    So I've heard ... I've never been able to :( ...so I got good at drawing v-t diagrams.
    not everywhere in the World it isn't :) I'm used to "s" referring to an arc-length and S being a surface area while d for displacement gets cumbersome with calculus.
    Indeed ... it is a good idea for students to get used to that too.

    I would have worked it like this:
    The v-t graph for this is a line starting at t=0, v=u and ending at t=T and v=0 (note: not given the final time - doesn't matter).
    Notice that the line has a negative slope (sloping downward to the right) and the area is a right-angled triangle.

    The slope of this graph is the acceleration: a=-u/T (slope = rise over run)
    The area under the graph is the displacement: s=(1/2)Tu (area = half base times height)

    2 equations and two unknowns - use the second to get an expression for T and substitute into the first:

    T = 2s/u (since you know s and u, most people make the mistake of computing T now. It's normally fine at this level but risks rounding-off errors later.)

    a = -u/(2s/u) = -u2/2s

    ... this is the kinematic equation that was used without the useless bits.
    It is faster to memorize the kinematic equations. However, that way means you have to memorize five specific equations and this way you need only remember basic geometry and how to draw graphs - both general skills applicable outside kinematics.
     
    Last edited: Sep 11, 2012
  14. Sep 12, 2012 #13
    ok that makes more sense bc iam just not physics oriented so that help me. thanks!
     
  15. Sep 12, 2012 #14
    I never even tried. Very long time ago, I memorized the obvious ## v = v_0 + at ## and the less obvious (it was given way before any elements of calculus were given) ## s = v_0t + \frac {at^2} 2 ## and got good at massaging those (plus the energy-related equations) into whatever shape I needed.

    And yes, it was ## s ## for displacement :)
     
  16. Sep 12, 2012 #15

    Simon Bridge

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    Those are the ones most students end up memorising. The first is off the slope of the v-t graph (though you may have got it via y=mx+c) ... you only need 2 equations and your second one has a lyrical, singsong, quality that makes it easy to memorise. The "obvious" second one is the area of the general v-t graph done by triangle + rectangle or as a trapesium:

    Trapesium: s=(v+u)T/2

    A lot of the people who have trouble with the equation approach respond to the graphical one.
     
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