Problem

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

07:20

Problem 16 Medium Difficulty

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).

Answer

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

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Video Transcript

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'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'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'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'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'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'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'm going to continue rewriting this so I'm looking for the limit As the time approaches eight. Mhm Okay. Of All of this, I'm going to continue to simplify now. I can't just plug eight in. I will get zero divided by zero, which doesn'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's going to be equal to the limit as time approaches eight. Okay. Yeah. Of this is fact terrible. I'm going to first of all factor the one half out. So when I do that, that'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'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't even need that anymore. I don'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'm going to insert a graph. Yeah, hopes that's not what I wanted to do, insert a graph. I'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'm sorry. I mean is the amount a little bit Okay. Yeah. Right. I'm just trying to take a moment, make sure that I'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'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'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

Michael C.

University of Southern California

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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

More Answers

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Calculus: Early Transcendentals

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Calculus: Early Transcendentals

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Catherine R.

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Kayleah T.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Calculus: Early Transcendentals

Catherine R.

Caleb E.

Kristen K.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Grace H.

Catherine R.

Heather Z.

Kayleah T.

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