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# The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.$v(t) = t^2 - 2t - 3$, $2 \le t \le 4$

## a) $\frac{2}{3}$b) 4

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given velocity function V f t which is equal to t squared minus two t minus three. We want to find the displacement and the distance traveled by the particle Over the interval 2- four. For the first part we have s this is equal to The integral from 2 to 4 of VFT which is just The integral from 2 to 4 of He squared -2 T -3 Bt. And integrating, we have 1 3rd TQ two times t squared over two minus three T. This evaluated from 2 to 4 and simplifying, we have 1 30 cube minus t squared minus three T. This evaluated from 2 to 4 and when these four we have 1/3 times four. Race to the third power minus four squared minus three times four. This minus 20 is too, we have one third times two to the third, power minus two squared minus three times two. This gives us the value of to over three and so the displacement of the particle Is to 3rd m for part B B house distance. This is equal to the Integral from 2 to 4 of the absolute value of VF t d T. And this is equal to the integral from 2 to 4 of the absolute value of t squared minus two t minus three Bt. Now the absolute value of t squared minus two t minus three. This is equal to the absolute value of t minus three Times T Plus one. This is equal to positive t minus three Time's D Plus one. If t is in The interval 3-4 And you have negative of T -3 time's D plus one yes T is in 2- three. And so we rewrite this into the Integral from 2 to 3 of the negative of t squared minus two t minus three D. T. Plus we have The integral from 3 to 4 of t squared minus two t minus three DT. Now integrating we have negative of one third t cube minus we have to t squared over two, that's D squared minus three T. This evaluated from 2 to 3 plus We have 1/3 t to the third power minus t squared minus three. T. This evaluated from 3 to 4. Now when these three we have negative of one third times three to the third power minus three squared minus three times three. This -20 is too you have one third times two to the third, power minus two Squared -3 times two. And then this plus we have when these four that's 1/3 Times 4 to the 3rd. Power minus four squared minus three times four. This minus when T is three we have 1/3 times 3 to the third power minus three squared minus. We have three times three and this gives us Negative of we have 9 -9 minus nine minus. We have 8/3 -4 -6 and then plus we have 64/3 -16 -12 minus. We have 9 -9 -9, and so this gives us the value of to over three, and so the Distance traveled by the particle is 2/3 m.

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