the-displacement-in-feet-of-a-particle-moving-in-a-straight-line-is-given-by-s-frac12t2-6t-23-where-

Question

The displacement (in feet) of a particle moving in a straight line is given by $ s = \frac{1}{2}t^2 - 6t + 23 $, where $ t $ is measured in seconds.

(a) Find the average velocity over each time interval:
        (i) $ [4, 8] $                 (ii) $ [6, 8] $ 
      (iii) $ [8, 10] $              (iv) $ [8, 12] $ 

(b) Find the instantaneous velocity when $ t = 8 $.

(c) Draw the graph of $ s $ as a function of $ t $ and draw the secant lines whose slopes are the average velocities in part (a). Then draw the tangent line whose slope is the instantaneous velocity in part (b).

Daniel Jaimes · Accepted Answer

This is problem number sixteen of the Stuart Calculus. Eighth edition, Section two point seven. The displacement and feet of a particle moving in a straight line is given by as equals one half a T squared minus sixty plus twenty three. Forty is measured in seconds. Party. Bring the average velocity over each tiny material forty sixty eight to ten. Eight to twelve. Ah, what would help us out is to find the values of ass for each of these times team. So just going to make a quick table If we plug in, pour into our equation. Fres, we should get seven as your answer. If we plug in six, we should have five and free play an eight. We should get seventy. Enter every play in ten. We should get thirteen. And finally, if we plug in twelve as t we should kept twenty three. And the reason these points are important is because we want to use specific average velocity equation. Um, which is function so in our case has ah, average velocity is s of he minus s all rights of t one t two minus as fifty one. Ahh lower TT misty one two It is the slope between two specific point generally. And if we do it for each of these points or each of these intervals, we should can&#x27;t for part one. We&#x27;re going to use their table and we&#x27;re gonna evaluate what&#x27;s toa? A function equal to eight. It is seven for seven minus the function evaluated at four, which is also seven divided by eight minus for ten minus seven zero. Divided by four is zero. So that&#x27;s the average velocity for part one. Part two, It&#x27;s eight minus six. So s of eight s of eight, which is seven. My missus of six, which is five divided by eight ninety six. This is seven minus five is to do it right. A minus six is too. She was an answered one. Now each of these are in feet per second said these air the average velocity zero feet per second For part one of the party one foot per second. Part two of party part three. We have the interval for me to take him so we get as attendance. Thirteen minus X rayed is seven divided by ten minus heat and this is dirty money seven to six stores for a ten minute status. Two, we get three feet per second and finally for part for we have the interval for me to twelve. So s twelve, which is twenty three minus a survey to which a seven divided by told, minus eight twenty three minutes seven sixteen hundred told me eight, which is for sixteen years before is for feet per second. All right, so we have calculated all the average velocities for each of the time intervals. We gets here a few per second for this term. Mineral report eight one foot per second for the timing. Around sixteen, three people second from the terminal for me. Ten, four and four feet per second for the attainment of all From it to talk part B. We&#x27;re looking for the instantaneous velocity for this function. And we will Uh huh. We use equation, too. For this, the instantaneous velocity is the limit as h purchase zero out of this function, evaluated at eight plus ity one half, eight plus team squared six times, Ade Close team plus twenty three. And this will be minus the function Evaluate eight, which we already calculated to be seven and then all of this to get by h My mistake. I will go back and replaces with age because it is supposed to be a plus age, not Abel&#x27;s team. This will be eight plus each eight bucks each king. Our next step is to simplify the numerator. We&#x27;re going to expand into the terms. One head of state. It&#x27;s for one of Tim&#x27;s. Um h squared is one all. First, we actually we have to expand this term because this is ah, binomial squared. So we foil. We should get sixty four plus sixteen h plus age squared. That&#x27;s if it&#x27;s sixty four. Sixty four, right, too. Thirty two on the next term of sixteen h good bye to his age. And then the last term is H squared. But the one half in the front minus six times Hate, which is forty eight minus six each plus twenty three minus seven. All that of our age. Okay, if we work with the constants just a Constance, we get thirty two minus forty eight is negative. Sixteen plus twenty three is positive. Seven minus seven zero. So all they can Constance will cancel out the remaining terms, each having each Let&#x27;s cancel our one age for each term in the numerator with the denominator each leaving us with, um the Lim. And each purchase zero eight minus six is two class, one half each. And if h approaches zero, they won&#x27;t have aged Turn vanishes and we&#x27;re left with just two. And that is our instantaneous velocity and the TV it is to fever. Second, this consistent with our work because eight between these three intervals here, specifically between six to eight and eight to ten and we see that the slope to why is between the two slopes the two average velocities here won a few per second and three percent finally part See, draw the graph of s as a function of team and draw the secret lines who slopes are the average velocities in party. So slip of zero. So one slip of three. So before then, draw the tangent line. Who slope is the instantaneous velocity in part B and then the team&#x27;s posit years too. So this is something that we will we can sketch. In this case, we used a plug in device and found each other functions. Although this isn&#x27;t necessary, we just need to to sketch from the sketch of the function as one have t squared minus sixty plus twenty three. Is this probably here in black? We uploaded all the points that were relevant for seven, six, five, eight, seven, ten, thirteen and twelve. Twenty three on we see the average velocity shown for each interval before the eight day average velocity was zero. So we see this flat lines horizontal line for the interval. From a sixty, we have this orange line which has won from eight to ten. We have this blue line with which has a slip of three for eight to twelve. It&#x27;s a much steeper line with this look before that&#x27;s in purple and the instantaneous velocity. The tension line is the one shown in red, and this definitely has a slope of two. So all of this is consistent, consistently showing that average velocities can closely s Timmy What? The slope is instantaneously a certain time, but it&#x27;s only when used the limit as h approaches here or as t approaches Teo point of interest in this case, eight that you are able to get exactly, uh, tension line slope, which is equal to to in this case

Hariprasad Annamalai · Answer

everyone. So what we have is the motion of a particle that is to find by this equation for position here such that the position is a function of time. Is it cubed? Minus 60 squared minus 15 T plus seven. What we&#x27;re asked to solve for is the total distance traveled, the average velocity, the average speed and the instantaneous velocity at most 10 seconds. So what we&#x27;re going to do first is go ahead and take our expression for s. Our position, which is to recall s, is equal to take minus 60 squared minus 15 T plus seven. And take the derivative of this. Just get our velocity. So we see that V is equal to De S DT, which gives us three D squared minus 12 T minus 15. So we can actually go ahead and solve for r instantaneous velocity here. So, by plugging in t equals 10 we&#x27;ll see that the V Atiq with 10 is 1 65 ft per second. So that&#x27;s one part of our answer. So we go ahead box that off next week and take the derivative of velocity to get acceleration and we can say that a is equal to DVD T, and this gives us that acceleration is a function of time. This 60 minus 12. Now we can plug in, uh, t equal to 10 and we&#x27;ll see the acceleration to equal to 10 is equal to 48 feet per second squared. This is another part of our answer. We could get hood and box. That a swell. Now, given that this motion is governed by a polynomial to solve for our total distance traveled, we need to solve for when the particle change direction and also find the positive route so we can see by solving this equation for zero is that a positive route occurs when t is equal to five seconds. So therefore, what we can say that AT T equals zero s is equal to 7 ft. That he equals five s is equal to negative 93 ft and at our final time, point of T equals 10. This is equal to 257 ft. So by summing these up, we can find the total distance that has traveled. So we see that this goes from positive to negative back positive. So what this means is that we can solve for the total distance. They&#x27;re saying 7 ft plus 93 ft times two because we&#x27;re going down and back to 93 coming back to zero plus 257 ft. This gives us a total distance, traveled equal to 450 ft. So this will be another part of our final answer. Now, we&#x27;re also asked us all for the average velocity, A T equals zero so called that average velocity is equal to Delta s all of adult T, and this will be equal to our displacement. So Delta Ass is 2 57 minus seven all over about the tea, which is 10. And this gives us average velocity 25 feet per se. And now we look at our average speed. And remember that average speed call it. The average is just total distance traveled all over Delta T. And this is equal to 450 all over 10 seconds, which gives us an average speed of 45 ft per second. And that is there a complete final answer

Heather Zimmers · Answer

here we have a position equation and for part a of the problem. We want to find the velocity. Remember that velocity is the derivative of position. And because our position equation is a quotient, we&#x27;re going to find the derivative using the quotient rule. So here we have the bottom times, the derivative of the top, minus the top times the derivative of the bottom over the bottom squared. Next, we need to simplify that. So what I&#x27;ve done is distribute the nine and multiply the 90 and the two t And then I combined the like terms. So that&#x27;s our velocity for part B. We want to find the velocity at time one. So we&#x27;re going to substitute a one in for tea and then simplify and we end up with 18 25th and the units would be feet per second for part C. We want to know when the particle is at rest and a particle is at rest if its velocity ease is equal to zero. So we take our velocity, we set it equal to zero, and we saw for time. Now, when you have a fraction that equal zero, you want the numerator to equal zero. And when you look at this denominator, you&#x27;ll notice that there&#x27;s no real number that makes it equal to zero anyway. So we set the numerator equal to zero, and we saw for tea we can add 90 square to both sides. Now we have 90 squared equals 81. We can divide both sides by nine. Now we have t squared equals nine and weaken square root. And typically we would write plus or minus three. But because for this problem we&#x27;re starting at time zero and only using positive times, we&#x27;re not going to include negative three. Okay, so now we know the particle is at rest at time. Three seconds for part D. We want to know when is the particle moving in the positive direction and it will be moving in the positive direction if the velocity is zero. So what we do is we take the information from the previous part. We found out that the velocity was zero when the time was three. And what we do then is we make test intervals before three and after three to figure out what the velocity is doing. So before three, we have the interval from 0 to 3 and after three, we have the interval from 3 to 0. So what we want to do is pick a number in each interval and test it into the velocity to see if we get a positive or negative. So I picked one, and I picked for and remember, back in the previous part of the problem, I think it was part B. We found that V of one was positive 18 25th. So we know that via one is positive, and then if you substitute for into the velocity, you get a negative. So what this is telling us is that the particle is moving in the positive direction from time 0 to 3, and it&#x27;s moving in the negative direction from time. Three to Infinity for party. We want to find the total distance traveled by the particle from time zero to time. Six. So here&#x27;s what we just learned. We learned that it&#x27;s traveling in the positive direction from time zero to time three and then it&#x27;s traveling in the negative direction from time 32 times six. So what we don&#x27;t want to have happen is we don&#x27;t want a negative to cancel out the positive distance. And so we&#x27;re going to take the absolute value of each distance. We&#x27;re gonna break it down into the distance from time zero to time three and the distance from time 32 times six and take the absolute value each time. So to get the distance from time zero to time. Three. What we do is we find the position at time three minus the position at time zero and then we take the absolute value and to get the distance from time 32 times six. We take the position at times six, minus the position a time three and then take the absolute value. Now those position values come from the original function F of X. So if you need to find F zero F of 3.5 of six, you can substitute those into the your F of X function, and you can use the table feature of your graphing calculator to do that. It&#x27;s really helpful, so those numbers go in. So we have the absolute value of 1.5 minus zero, plus the absolute valuable in 00.2 minus 1.5, and that gives us the absolute value of 1.5 plus the absolute value of negative 0.3. And so we have 1.5 plus 0.3. So the distance travelled is 1.8 feet. Now we&#x27;re going to make a motion diagram and because we know it went in the positive direction for a while and in the negative direction for a while, we should see that in our motion diagram. So what we see is a particle starting at time zero position zero. It moves to the right and it stops moving to the right a time three. And then it moves to the left for part G. We want to find the acceleration. And remember, acceleration is the derivative of velocity. So we had our velocity equation. Now we take the derivative of it using the quotient rule, and we get our acceleration equation. So here we have the bottom times, the derivative of the top minus the top times, the derivative of the bottom, using the chain rule to do that over the bottom squared, that&#x27;s going to need to be simplified if we&#x27;re going to be able to use it for anything So I distributed and multiplied these together. Actually, no, I did not distribute. What I did was I factored out a t squared plus nine from here and a T squared plus nine from here. And I canceled it with one of the t squared plus nine on the bottom. So that left me with what we see in the next step. And then I distributed. So I multiplied the 18 t by t squared plus nine and I multiplied the fourty by 19 minus 90 squared. And once we have that, we can combine the like terms. And then I decided after that that I would factor out the common factor of the numerator which was 18 t because I&#x27;m anticipating that eventually we&#x27;re going to get to the point where we have to set this equal to zero. And I thought it would be easier if it was factored. Okay, The next thing we want to do is find the acceleration a time one. So we substitute one into that acceleration function and we can simplify it and we get negative 1 17/2 50 the units would be feet per second squared. Now that we have the position, the velocity and the acceleration. We can use a graphing calculator and look at all three graphs. So we go to the calculator, we go to the y equals menu, we type them all in, and now we need to set a good viewing window that&#x27;s going to allow us to see all of these. So think back over some of the numbers we got from the previous parts of the problem. Like we found out that the positions were numbers like zero on 1.5 and 1.2. We found out that times were numbers like zero and three and six. So based on that, I went to a window and I decided to try letting my ex values go from the negative 1 to 10 and letting my Y values go from negative 2 to 3. And you can also always fiddle with these numbers and make changes until you like what you see. So what we have here, the blue one is the graph of the position equation. The red one is the graph of the glossy equation on the black one is the graph of the acceleration equation. Okay, After that The final part is to find out when the particle is speeding up and when it&#x27;s slowing down and a particle will be speeding up if its velocity and acceleration are both positive or both negative and it will be slowing down if its velocity and acceleration have opposite signs, one with a positive one with a negative. So we need to figure out when our acceleration is zero so that we can then figure out when it&#x27;s positive and when it&#x27;s negative. So we take our acceleration and set it equal to zero, and you&#x27;ll notice here that I just have the numerator because if a fraction is equal to zero, it&#x27;s just the numerator that&#x27;s going to equal zero. And so then we can set each factor equal to zero, and we can solve each of those and we end up with T equals zero or T equals three square root three. And we&#x27;re not including the negative time, of course, because we&#x27;re only talking about times starting at zero and going forward. Okay, so what we&#x27;re going to do once we know the times that the acceleration is equal to zero is break the number line down and we&#x27;re going to find the acceleration between time zero and time three Square root three and we&#x27;ll find the acceleration between time three square three on infinity. So what I did was I plugged in a number into the acceleration. I plugged in a number that was in the first interval. We had already plugged in a one, so that&#x27;s convenient and it turned out negative. And then I took 10. I never That was in the second interval. Put it into the acceleration and I got a positive. So now let&#x27;s let&#x27;s bring that information together with the information from the velocity. So I&#x27;m breaking down the intervals and I&#x27;ve broken down the intervals at three square Root three because that is where we see the change in acceleration. But I&#x27;ve also broken down the intervals at three because that&#x27;s where we saw the change in velocity. So what we saw with velocity was it was positive between zero and three, and it was negative after that. What we saw with acceleration was that it was negative before three square or three, and it was positive after that. So based on that, when the signs are both the same negative and negative. We know the particle is speeding up, so it&#x27;s speeding up from times 3 to 3 square root three. And when the signs are different, a positive and a negative, we know that the particle is slowing down, so it&#x27;s slowing down between times zero and three and times three, square three and infinity.

Mike Gaerlan · Answer

Okay, so we have a particle that is moving along a path, given this given this time here. So we need to find the average velocity between t equals one and one plus h. All right. First part here, if H is equal to 0.1, and so average velocity is going to be s at T the final T, which is as one plus h eso. Go ahead and put in 0.1 here. Um, minus s at one divided by, uh, this, uh, t here, which is 1.1 minus one. So this is going to give us all right? So if we plug in 1.1 into s here. All right, so we just go ahead and plug that into account stated, um, that&#x27;s 41.1 squared plus three, then minus. Um, for times one squared minus. We&#x27;re starting plus three all over the bottom part. There is going to be 0.1. Okay, again. So this top part here, people. Dental calculator. That is 7.84 Linus. This part here is seven. Divided by coins one. This gives us 0.84 divided by 0.1. Okay, And then now, uh, we have, um that our final average velocity, it&#x27;s just going to be 8.4. And our units here are in meters for a second. Ass is in meters and T is in seconds. Right? So that we have 8.4 meters per second. Now do the scene. Okay for H is equal to 0.1 01 And so we&#x27;re gonna have s of 1.1 minus s of one. Divided by 1.1 minus one. Okay, so I&#x27;m gonna go ahead and put this in to a calculator. If I play that include calculator. I get, um, seven point 00 our sorry. 7.80 for minus seven, divided by 0.1 So then I get 0.0 804 divided by 0.1 and I get approximately 8.4 meters per second. All right, now, lastly, we&#x27;re going to do H equals 0.3 Okay, You&#x27;re or saris. Airplane 00 one. So again, so s at 1.1 minus s of one. Divided by 1.1 minus one. If you plug that into our confidence now, Okay, we&#x27;re going to get 7.0 um, 7.8 004 minus seven. But by 0.1 what&#x27;s we&#x27;re gonna get to your 0.8004 but by 0.1 This gives us 8.4 leaders for second. So if you can see a pattern here, um, well, notice We&#x27;re getting closer and closer to eight. Right? This vessel part here is just gonna get smaller and smaller and smaller and smaller as we get to H. All right, so we can say that our estimate for, um so that&#x27;s part A. Our estimate for our instantaneous velocity will be about eight meters per second.

Keshav Singh · Answer

for this problem on the topic of cosmetics of particles were given the emotion off a particle during a 14 2nd journey. It has constant acceleration for its first part of the journey, zero acceleration for the next part and again, constant acceleration. For its final part, we know the time for each part of the journey as well as its velocity. During the zero acceleration phase, we are asked to construct the V, T and X curves for the total journey, as well as the position and velocity of the particle and its total distance traveled after he total time has elapsed so well. First, construct the acceleration time curve and we&#x27;ll label the area from 0 to 6 seconds. We&#x27;ll call it a one, and that&#x27;s minus 24 ft per second and the area under the curve from 10 to 14 seconds. The final constant acceleration phase, which is 16 ft a second. So labeling it a one and a two. We have a one two B minus 24 ft second, and the area under the exploration time graph for the second constant acceleration phase is 16 feet a second, and you can see that the units often accept off a velocity. So let&#x27;s look at part A. And first we need to construct the VT cough. Now we know that at six seconds the six is given to a minus eight feet a second, and so we can calculate the initial velocity. V, not a T is equal to zero. This is simply V six minus a one, the area under the 80 graph for the first part of the journey. This is minus eight minus minus 24 which gives an initial speed off 16 ft a second. Now, at 10 seconds, we have eaten to be again minus 80 ft per second since this previous constant from 6 to 10 seconds. And the final velocity V 14 is equal to whatever the speedy&#x27;s at time 10 plus the area under the acceleration time curve from 10 to 14 seconds. Remember, the area on an acceleration time curve gives a D velocity, so that&#x27;s minus eight feet per second. Class 16, which gives the speed at 14 seconds to be 8 ft the second. And so from here we can sketch the VT curve as follows and from the Bt curve, we label the areas. A three, a four, a five, a six and a seven, as we have in the diagram. So these areas the areas under DVT cough give us position. That&#x27;s can calculate these areas from the curve. So we firstly have a three, which is the area of a triangle to be 32 feet. The area a full two B, minus 8 ft a five is minus 32 ft. A six is minus 8 ft and a seven final area is equal to 8 ft. So now we have the areas and will use this to calculate or to construct the exchequer brother. Okay, now to construct our X T curve. Yeah, we&#x27;ll do this as follows. We have the initial position. It&#x27;s not to be zero. He positioned after t is equal to four seconds. X four is equal to x note. Lastly, area a three. And so this is simply 32 feed. Since the initial position of zero he positioned at teasing of six seconds at six is equal to X four plus area a fort and this is 24 ft. X 10 is equal to x six plus eight five. It&#x27;s and this is minus 8 ft after 12 seconds. The position of the particle X 12 is equal to x 10 Okay plus area in six under the meeting draft, which gives us position x 12 to B minus 16 ft and finally the final position x 14 at T 0. 14 seconds is x 12. So last area A seven, which is minus eight feet. Now, with this information, we can plot the curve as follows. And so from hour X T curve has shown we can calculate the total distance traveled. Okay, so the total distance traveled. We&#x27;ll do this in sections from 024 seconds. We have distance traveled. The one is equal to the absolute value off 32 minus zero, which is a total distance off 32 feet between times four seconds and 12 seconds. We have a distance D two, which is the absolute value off minus 16 minus 32 ft, which gives us a total distance traveled or 48 feet, and between a time of 12 seconds and 14 seconds. We have a distance. The three, which is the absolute value off minus eight minus minus 16, which gives us a final distance in the last two seconds off 8 ft and therefore the total distance traveled. We&#x27;ll call it D is 32 last 48 last eight, which is a total distance that the particle travels off 88 ft.

Zulfiqar Ali · Answer

Well, position is a function of time. Could be regiones integral 40 times. DT So, uh, this into girl is equal to three d squared minus 60 times DT. Since we equals Treaty Square minus 60 and solving this integral, our position is a function of time Musical two D, Q Minus free T square. Let&#x27;s see where sees a constant well, it Time t is equal to zero. Uh, position is zero Q minus three to CDO square place scene and therefore si is equal to four. So C is equal to four, right? Since ah c equals all right position in Time T is equal to zero, So seize equal to four and further, like music rollup um, we have position is a function of time is equal to a cube minus three t square less four feet and position when time physical four seconds is equal to forgive Minus three into four square this fuller musical to 20 peach right, So we have that&#x27;s not is equal to four feet and s one is equal to 20 feet and some office not And this one is equal to 24 feet. No, let&#x27;s find out acts relation we don&#x27;t which is, um TV divided Fine TT eyes equal to x relation of the function of time which is equal to 60 minus six feet or second squared. So x relation when time physical two seconds His six into two mine six Matiz ical to six beat were second square.

Daniel Jaimes · Answer

Okay. Hello? So here were given the displacement of a particle moving in a straight line is given by this function here S. F. T. Is equal to one half a. T squared minus 60 plus 23. So for part A for the first part here we have the party we had the average velocity of the particle over the time in about a baby. That&#x27;s going to be S of b minus S. Of A. All divided by b minus eight. So for the first part we have our interval from 4 to 8. Therefore we have, the average velocity is S of eight minus s. Of four, all divided by eight minus four. So that&#x27;s going to be one half times eight squared minus six times eight plus 23. That gives us 30 to -48 plus 23. And then we have um minus well one half times four squared minus six times four plus 23. That gives us an eight minus 24 plus 23. And that is all divided by eight minus four. Is are all divided by four. Well this just gives us this first part is seven. The second part is seven. Therefore that&#x27;s just seven minus seven, 7 -7 or 797 All over four. So that&#x27;s equal to zero. So therefore the average velocity here is equal to zero ft per second. Yeah. All right. So there&#x27;s our first part for the second part um We have the individual from 6 to 8. So for part two of part A we have well one half times eight squared minus six times eight plus 23. That&#x27;s going to be while 30 to minus 48 plus 23 minus 18 9 36 plus 23/2. Which gives us to over two which is going to give us one. So therefore Part Two here um The average velocity is one one ft per second or one ft/s. And then for the third part we have the interview from 8-10. So for the third part here from eight we have again there&#x27;s going to be s of 10 minus s of eight, All divided by 10 -8. And now we get um while 50 minus 60 plus 23 minus the quantity, 30 to minus 48 plus 23 all over two and then about two weeks to three. So this third part three here we get three three ft/s. And then for part four we have the interval from 8 to 12. So if a part four we have S of 12 minus S of eight, all divided by 12 minus eight. And that&#x27;s going to be um evaluate to end up being four. So the value here is gonna be four ft per second. Alright then going on to Part B. So part B. The objective here is to find the instantaneous velocity when T. Is equal to eight. So we&#x27;re going to have here V. of eight is going to be equal to well the limit as as a church approaches zero of S. Of um eight plus H minus S. Of eight. All divided by H. So here um we end up with the limits can be the limit as H approaches zero of why don&#x27;t we simplify this? This just becomes a one half H squared plus two. H. I&#x27;m all over it. Which the H is canceled. This is going to be equal to the limit as a church approaches zero of um just one half H plus two. And going ahead and taking the limit this term here goes to zero. So therefore the limit here is equal to two. So therefore the instantaneous velocity when T. Is equal to H is going to be too Uh two ft/s, two ft/s is our instantaneous velocity there. Um And then going on to part C. Where all the average velocities in part represents the slope of the C. Can&#x27;t lines. So um right, we could go ahead and graph this function. Um And we get S of T. Is going to be equal to one half a t squared minus 60 plus 23. Yeah. And the we get sort of the graph of that is going to be a parabola that looks something like. So, So we have let&#x27;s see here at the uh at six is going to be our vertex. Um We&#x27;re going to look did you coming up at four? So maybe this is again not going to be the scale, but we get here, we have our Parabola that looks something like, so we come up here and why access is going to be up at 22. And then we have um the tangent line, which is going to be look something. Maybe we&#x27;ll put in red um looking like so Okay.

Daniel Jaimes · Answer

in this activity were given that the displacement of a particle moving in a line is given by this function one half a t squared minus 60 plus 23 where t. Is time in seconds. And we&#x27;re being asked to find the average velocity at several different intervals over seven. Several different intervals time. 1st from four seconds to eight seconds. The average velocity of this particle between four seconds and eight seconds. Well, average velocity is given to me as the change in displacement divided by right. The change in time. Yeah. So I want to calculate uh huh The change in time. And I want to calculate the displacement at the two different times and subtract them to get the change and displacement at each time. And then that will give me my average velocity mm I&#x27;m going to set this up so that I can calculate displacement. The average velocity between four and 8. Let me just draw this instead of typing it. The average of displacement between four and 8 is going to be The displacement at eight. So when time equals eight displacement equals five, seven minus the displacement. Have four seconds. The displacement at four seconds also equal seven, divided by the change in time, which is eight minus four, seven minus 708 minus four is 40, divided by four is zero. So the displacement between four seconds and eight seconds. Zero. Then the displacement between six seconds and eight seconds. Is it going to be given to me? The average velocity will be given to me by the displacement at eight seconds. Which we already know is seven minus the displacement at six seconds. The displacement at six seconds, soups is five Over the change in time which is six seconds -4 seconds. 7 -5 is too 6 -4 is too Two divided by two is 1. So the average velocity over that time Interpol is one. Then The time interval eight seconds to 10 seconds velocity is given to me by the change in time Or the change in displacement divided by the change in time to the displacement of this particle at 10 seconds is 13. And this displacement at eight seconds we already know is seven divided by 10 -8. 13 -7 is six. 10 minus eight is to Six point is too is threes of the average of velocity is three ft per second over that interval. And then finally Where has to calculate the displacement from eight to 12 seconds. Yeah. So the average velocity Is going to be the displacement at 12 seconds Which is 23 minus this minus the displacement at eight seconds. Which is seven Divided by 12 -8. Yeah, 23 -7 is 16. -8 Is 4. 16 divided by four is 4. Okay. Yeah, Yeah. So the average velocity between eight seconds and 12 seconds is four ft per second. That&#x27;s are part a For part being we are being asked to find the instantaneous velocity at time equals eight seconds. So I am going to go ahead and erase this that I&#x27;ve got room to work on that. So we remember that for part paying The velocity is Part one was 0. The velocity on the second time interval was one capacity On 3rd time interval was three. The velocity on the 4th interval was four. Then for part B. The instantaneous velocity is given to me as the limit of the average velocity as some time gets closer and closer to the input that I&#x27;m interested in. Eight. So my input time is getting closer to eight seconds and I am finding what is the limit of the average velocity. As the difference between the time and eight seconds itself gets smaller and smaller. So that is the displacement at time T Minour the displacement at time eight divided by t minus eight. That&#x27;s the average velocity. If I plug in the formula for the displacement, that becomes what is the limit. Yeah. As my time T gets closer and closer to eight, Who Not Infinity eight. Yeah. Uh Yeah. My displacement at time T is 1/2 of T squared minus 60 plus 23. My displacement at time eight is subtracted and that is minus one half of eight squared which is 64 Plus six times 8, which is 48 plus or minus positive, 23 minus 23. All divided by my input. Time t minus eight. The difference between time T and eight seconds. Well, first thing that I notice is 23 minus 23 0. Then I&#x27;m going to continue rewriting this so I&#x27;m looking for the limit As the time approaches eight. Mhm Okay. Of All of this, I&#x27;m going to continue to simplify now. I can&#x27;t just plug eight in. I will get zero divided by zero, which doesn&#x27;t help me at all because T -8 would become 8 -8. That is undefined. Mm So I am going to see if I can simplify this and factor it all got one half a T squared uh minus 60. Okay well what? Yes, okay minus 16 -1 half of 64. His 32 plus 48. All divided by T -8 and 48 -32 is 16. So Change this to a positive 16. Yeah. Mhm. That&#x27;s going to be equal to the limit as time approaches eight. Okay. Yeah. Of this is fact terrible. I&#x27;m going to first of all factor the one half out. So when I do that, that&#x27;s T squared minus 12. T plus 32 Divided by T -8. And if this is factory bubble, so this all becomes the limit As T Approaches eight. Okay. Of one half of T -4 times t minus eight. Mhm Divided by T -8. Now here&#x27;s the magic part, T -8 cancels out the T -8 after I factored it And this becomes just the limit As T Approaches eight. Yeah. Are you? Yeah. Yeah. Of well I don&#x27;t even need that anymore. I don&#x27;t need my denominator because I just canceled it out. So the limit As he approaches eight of 1/2 of T -4. And that I can just plug the eight and eight minus four is 41 half of four is too. So the instantaneous velocity for part B, the instantaneous velocity. Hat Okay, seconds is to feet per second. Then the last part of this problem is to graph the parabola and graph the sequence that represent those average velocities. And graph the line the tangent line that represents that instantaneous velocity. So I&#x27;m going to insert a graph. Yeah, hopes that&#x27;s not what I wanted to do, insert a graph. I&#x27;m going to move it here so that we can make it bigger. Yeah, graphing are Parabola, you can do this with technology. Mhm. Okay. Yeah. Mhm. I&#x27;m sorry. I mean is the amount a little bit Okay. Yeah. Right. I&#x27;m just trying to take a moment, make sure that I&#x27;ve got a good scale since I only need to go up to 12. Okay, so sketching these sequence the 1st 2nd was between or and X. Yeah. Can you recall that for and at eight? The velocity was seven. So we had an average rate, An average velocity of zero because the displacement went from seven, it moved back and then it went back to seven. So the overall change was zero. Then we had The second between six and eight. So that line represents the average velocity between six and 8. Then we had a secret between eight and 10. That secret represents the average velocity between eight seconds and 10 seconds. And then we had one between eight and 12 and I didn&#x27;t make this quite big enough, but between A and 12 which is going to be up here, the second would represent that average velocity. So those are the three. Average velocity is represented as the sequence between the um ordered pairs of the displacement versus time. And then we&#x27;re asked to graph the tangent line that represents the instantaneous velocity at eight seconds. Eight seconds is right here. The instantaneous velocity was too. So that line is going to have a slope of two. Which means that the tangent line would look kind of like this. Okay. Oh

Problem

For the function $ g $ whose graph is given, arra…

Problem 16 Medium Difficulty

Answer

a) see solution
b) 2 $\mathrm{ft} / \mathrm{s}$
c) see graph

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James Stewart

Calculus

Kayleah T.

Caleb E.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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a) see solutionb) 2 $\mathrm{ft} / \mathrm{s}$c) see graph

Calculus

Grace H.

Kayleah T.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartCalculus

Kayleah T.

Caleb E.

Kristen K.

Samuel H.

a) see solution
b) 2 $\mathrm{ft} / \mathrm{s}$
c) see graph

James Stewart

Calculus