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I have to find the distance travelled by a particle between t=0 and t=4 secs for x(t)= 4t^2 - 2t^3. I tried differentiating the equation to find v(t) then integrating it within t=0 and t=4, but that got me the displacement. I'm pretty stumped. The maxima is at t=4/3, which is when the velocity turns negative.

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2 Answers 2

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Understand what is actually happening - the body first moves with some speed in positive x-axis direction, and at time = 4/3 the velocity becomes 0 and the body reverses its direction of motion. The thing which you are calling maxima, it is the point where the rate of change of function with respect to time is 0 - which means velocity is 0. Before and after that critical point, the sign of derivative is different - which is the direction of velocity is different.

To find the distance basically decompose the motion into two parts - one, where it is moving towards positive x axis and second, when its moving towards negative x axis. Integrate the function breaking the motion into these two parts. Once integrate it between t=0 to t=4/3, second integrate it between t=4/3 to t=4, the answers you get in both - add the absolute value of them to get your answer.

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The distance travelled is just x(4s)-x(0s) so it is just x(4s), since usually distance means the absolute value, you have to take |x(4s)| Since there are no units, one does not know if its miles or m or???

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  • $\begingroup$ ah, sorry. It's metres. The issue is that taking the value of |x(4s)| does not give me the right answer, as it gives the position of the particle at t=4s, which is not the same as the distance travelled by the particle in 4 seconds. $\endgroup$
    – plado
    Commented Sep 4, 2022 at 14:33

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