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Maths-displacement graphs

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???
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.
(edited 6 years ago)
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?
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)=dxdtv(t) = \dfrac{dx}{dt} graph and the time axis on a velocity-time graph.

So, you need to consider v(t).dt\displaystyle \int v(t) .dt (which you don't need to evaluate), but note that v(t)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 D=45v(t).dt04v(t).dt=[x(t)]45[x(t)]04\displaystyle D=\int_4^5 v(t) .dt - \int_0^4 v(t) .dt = [x(t)]_4^5 - [x(t)]_0^4
(edited 6 years ago)
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.
Original post by BuryMathsTutor
That would be true for a velocity time graph.


...

Long day, I suppose
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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