The velocity function is v(t) = t^2 - 6 t + 8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-1,5]. what is the displacement
Uh don't you have some sort of initial condition in the question? Otherwise when you integrate how are you going to solve for the constant?
well actually question is all about this nothing more actually i tried to solve this question but my supervisor said that isnot correct this is the solution that i answer to him -[-125/3 +75 - 40] - [-1/3 +3-8]
it's the integral of your function from -1 to 5. there's no need for an initial condition because it's asking for the distance travled IN the time span. \[\int\limits_{-1}^{5} (t ^{2} - 6t + 8) dt\]
lol oh yeah oops forgot
I think the question is actually a little more subtle that that. In this interval (unless I am imagining it wrong) the direction changes (twice!), so you need to conider this for the DISTANCE. The displacement is just your standard plug and chug integral.
yeah it crosses the axis at 2 and 4 so you just calculate the integral from -1 to 2, the integral from 2 to 4 and and from 4 to 5 and add all their absolute values up together.
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