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    The graph shows the resultant force (in the X axis) acting over an object with mass = 3 kg.


    The question is: When is the kinetic energy of the object at its highest level? I.e. when does the object have higher velocity?

    Thanks.

    PS: the graph is in Spanish, I know. :p:
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    There is a positive force being applied to the object for the whole 3 seconds. Therefore, the object is constantly accelerating according to Newton's Second Law, F(t) = ma(t) . Therefore, it's highest velocity is after 3 seconds.
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    (Original post by Phugoid)
    There is a positive force being applied to the object for the whole 3 seconds. Therefore, the object is constantly accelerating according to Newton's Second Law, F(t) = ma(t) . Therefore, it's highest velocity is after 3 seconds.
    Doesn't it start to decelerate at t=2?
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    (Original post by Miss Mary)
    Doesn't it start to decelerate at t=2?
    Nope. The magnitude of the force starts to reduce from 4 to 0, getting smaller and smaller, but it never goes negative in direction.

    A body only decelerates if the force acting on it is negative, but in this case, the force acting is always positive, therefore it is always accelerating, right until the force is equal to 0, at which point it maintains a constant velocity.
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    (Original post by Miss Mary)
    Doesn't it start to decelerate at t=2?
    No, the resultant force starts to decrease at t=2 which results in a decrease in the RATE at which the object is accelerating, however the object is still accelerating and gaining velocity.
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    According to impulse theorem F\Delta t=m\Delta v
    F\Delta t=\displaystyle \int_0^3\ F(t)dt= area of the trapezium =[2+(2+3)]\times 4\times 0.5=14Ns
    so m v_3-m v_0=14Ns \Rightarrow m v_3=mv_0+14Ns
    The object kept accelerating between 0s to 3s.
    When are the highest kinetic energy and higher velocity depend on v_0
    If v_0>0,of course both are at 3s.
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    (Original post by ATIRadeon9800xt)
    According to impulse theorem F\Delta t=m\Delta v
    F\Delta t=\displaystyle \int_0^3\ F(t)dt= area of the trapezium =[2+(2+3)]\times 4\times 0.5=14Ns
    so m v_3-m v_0=14Ns \Rightarrow m v_3=mv_0+14Ns
    The object kept accelerating between 0s to 3s.
    When are the highest kinetic energy and higher velocity depend on v_0
    If v_0>0,of course both are at 3s.
    Do you really need to make it that complicated? =P
    Surely just a logical thought process just comes to the same conlusion and is much easier to explain.
 
 
 
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