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    The displacement, x m, of a particle from a fixed point O at time t s is given by
    x= t^4 -4(t)^3-8(t)^2+1
    Find the total distance travelled by the particle during the first 5 seconds of motion???
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    (Original post by jasperkearns)
    The displacement, x m, of a particle from a fixed point O at time t s is given by
    x= t^4 -4(t)^3-8(t)^2+1
    Find the total distance travelled by the particle during the first 5 seconds of motion???
    Total distance travelled is the total area between the velocity graph and the time axis, when plotted against velocity-time axes.
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    (Original post by RDKGames)
    Total distance travelled is the total area between the displacement graph and the time axis, when plotted against time-displacement axes.
    Ok thank you very much, so just integrate then?
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    (Original post by jasperkearns)
    Ok thank you very much, so just integrate then?
    OK, ignore that, I meant to say that you're looking for the total area between v(t) = \dfrac{dx}{dt} graph and the time axis on a velocity-time graph.

    So, you need to consider \displaystyle \int v(t) .dt (which you don't need to evaluate), but note that v(t) is -ve between t=0 and t=4 and positive between t=4 and t=5 therefore you need to make your area between t=0 and t=4 +ve before adding it onto the area from t=4 to t=5

    So, the total distance traveled is just \displaystyle D=\int_4^5 v(t) .dt - \int_0^4 v(t) .dt = [x(t)]_4^5 - [x(t)]_0^4
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    (Original post by RDKGames)
    Total distance travelled is the total area between the displacement graph and the time axis, when plotted against time-displacement axes.
    That would be true for a velocity time graph.
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    (Original post by BuryMathsTutor)
    That would be true for a velocity time graph.
    ...

    Long day, I suppose
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    (Original post by RDKGames)
    ...

    Long day, I suppose
    I was struggling to believe that you had made a mistake, but we all do I suppose.
 
 
 

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