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alright now let's tackle Part B find the average velocity of the particle for the time period from zero is less than or equal to T is less than or equal to 6 so our average velocity that's just going to be our change in position which we could view as our displacement divided by our change in time well what is our change in position going to be well they don't give us our position function but they do give us our velocity function and so to figure out the average velocity we could figure out our displacement which is going to be equal to the integral of our velocity the integral of V of T DT and we want to find the average velocity for this time period so we're going to go T equals zero to T equals six and then we're going to divide that by the amount of time that goes by well our change in time is going to be equal to 6 and so this is going to be equal to the integral from zero to 6 of 2 sine of e to the T over 4 power plus 1 DT and then all of that divided by 6 where did I get this from well they tell us what our velocity as a function of time is it's that right over there so we can get our calculator out to solve this part so on our calculator we would hit math and then we would want to do number 9 which is definite integral so I hit 9 and we are going to go from 0 to 6 of this is going to be 2 sine of e to the so let me do second e to the and I'll just use the variable X instead of T because it's easier to type in X divided by 4 power so that's my sign and then let me close the parentheses for the sine plus 1 and then I'm integrating here with respect to X just because it was more convenient and I am going to get that and then I divide by 6 divided by 6 and I get and when you're taking an AP calculus exam it's important to round to three decimal places unless they tell you otherwise that's what they expect so it's approximately equal to one point nine four nine approximately one point nine four nine now let's do Part C so Part C find the total distance traveled by the particle from time T equals zero to T equals six so you might say hey didn't we just figure that out no this is displacement and to remember the difference between distance and displacement if I have something that starts here it goes over there and then it comes back to right over there the distance traveled is the total path length so it would be that Plus this while the displacement would just be this right over here so if you want to figure out distance what you want to do is take the integral of the absolute value of velocity you could think about it as you're taking the integral of the speed function so this is just going to be the integral from zero to six of the absolute value of V of T DT and then we can type this actually let me just write it out since you would want to do this if you're actually taking the test so it's integral from zero to six of the absolute value of two sign of e to the T over four plus one close the absolute value DT which is going to be approximately equal to so once again we hit math definite integral we're going from zero to six of now we'll hit math again and then we'll go to number we see absolute value the absolute value of 2 sine of e to the I'll use X again just because it's easier to use on the calculator and then I'll close the parentheses on whoops let me make sure I do this right so let me close the parentheses on the sine and then I have plus one and then there you go and then I'm integrating with respect to X approximately twelve point five seven three now let's do Part D I'll do it right over here Part D for T between zero and six the particle changes Direction exactly once find the position of the particle at that time so the way and I'm going to tackle it is first I need to think about at what time does the particle change direction and then I can figure out its position by taking the integral of the velocity function so how do I think about where that particle changes direction well you're going to change direction if your velocity is at zero and right before that your velocity was positive and then right after that your velocity is negative or the other way around right before that your faucet was negative and then your velocity after right after that is positive so we just have to figure out when does V of T equals zero so let's just say to sign of e to the T over four plus one is equal to zero and there's multiple ways you can solve this but because we're allowed to use a calculator we might as well do that so we hit math and then we hit B right over there I'll just hit enter equation solver so I could just say look the left side of my equation is 2 sine of we are going to go e to the I'll just use X as what I'm solving for because it's easier to type in in the calculator and close that parentheses plus one the right-hand side is equal to zero and now I just have to give an initial guess and I'll just go right in between that interval between T equals zero and T equals six so I'll just put a three right over there and now I just need to solve it and I can do that by hitting alpha and then solve and then you see it got us to five point one nine six so T is approximately equal to five point one nine six but remember they're not asking us for at what time does the particle change direction they want to know what's the position of the particle at that time well we know how to find our change in position our change in position is going to be our displacement from time equals zero to I'm 5.19 six and the displacement we're integrating the velocity function so that's going to be 2 sine of e to the T over 4 plus 1 DT now this is going to be our change in position but where did we start well they say at x equals 0 or when X 1 at time 0 our position was 2 so this is our change we would start here this would give us our total position where we started plus our change and so this is going to be approximately equal to we are going to have 2 plus and then I go to math definite integral from 0 to 5 point one nine six and I've typed this in many times already to sign of e of e to the X divided by four power and then close that parenthesis plus 1 and then I'm integrating with respect to X and so this gives us 14 point 1 3 5 approximately 14 point 1 3 5 is this position and let me make sure that I put those parentheses there in the right place and we are done

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