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Problem 62 Hard Difficulty
The acceleration function (in $ m/s^2 $) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time $ t $ and (b) the distance traveled during the given time interval.
$ a(t) = 2t + 3 $, $ v(0) = -4 $, $ 0 \le t \le 3 $
Answer
a) $t^{2}+3 t-4$
b) $\frac{-1}{3}-\frac{3}{2}+4 +9+\frac{21}{2}-12 -\frac{1}{3}-\frac{3}{2}+4$
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Top Calculus 1 / AB Educators
Catherine R.
Missouri State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
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Video Transcript
Alright I think give us the acceleration is to T plus three to t plus three then. Uh And we also know that the derivative of velocity is equal to acceleration. So if I find the integral of that uh it will give me the velocity so that now I. M. V. Of T. And the anti directive we add one to the expo divided by your new X 12 by by two is one Plus three T Plus C. And that's where we're going to use B. Uh Zero is equal to negative four because as I plug in negative four and for velocity and plug in zero for these two T 00 squared plus three times zero is just zero. Your left was C. Is equal to negative four. So the equation of velocity would be T squared plus three t minus four. Now as far as total distance traveled goes um Probably what I would do is first of all we care about from 0 to 3 zero 23 Um as I would see if there's any change in direction and you find the changing direction by setting this equal to zero. If you find factors in mega for that had to be three before -1. So there will be T plus four t minus one. So this object changes directions at T equals negative for which doesn't really make any sense and T.E. is one. Um Let's see. So what would I do? I would you're gonna have to find the integral of this, which is why this problem is so complicated to find that's right T D. T. The total distance traveled. That integral from 0 to 1 of that function. GT Yeah. Starting here and then it changes directions. Uh and then the integral from 1 to 3 of t squared plus three t minus four. Yeah. Gt uh Now one of these answers to be negative because we're moving to the left I don't know which one that is. I'm just doing the anti derivative From 0 to wine and then the same thing here really all the same work but as you plug in one you'll get one third plus three halves minus four. Yeah that's going to be a negative number. So absolute value that thing because total additions traveler needs to be positive and then add to it. This I guess I didn't show plugging in zero but you get a bunch of zeros. Um One third plus three halves one third times three squared cubed times three halves times three squared minus four times three. You can see why there's so much work to be done here minus plugging in all those ones now. On third plus three hands minus four. Yeah. Yeah. Uh I'll go ahead and put absolute value there just in case. Um And let's see yeah we can go to calculate and just type that in. Mhm. See absolute value that and then we squared Plus three cubed divided by two minus slopes. I'm using a calculator and I'm struggling right now minus 12 minus the quantity. One third plus 3/2 -4 hopefully attacked us all incorrectly because I don't want to go back and change anything. Um And I can change that to a decimal or to a fraction as well. Math track will say 89 6. Here we go.
Top Calculus 1 / AB Educators
Grace H.
Numerade Educator
Catherine R.
Missouri State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
