The displacement in feet of a particle moving in a straight...

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).

Transcript

00:01 Here's an equation for a particles movement at some time t, as tells you where it's located.

00:08 So they want us to find some average velocity, so i went ahead and calculated the position at those times.

00:15 So after four seconds, it's located at seven.

00:18 Like say it's on a ruler, it's on seven, and then after six seconds, it's at five, then it's back to seven, and then to 13, and then to 23.

00:27 Okay, part a.

00:28 Find the average velocity on these intervals.

00:31 So average velocity is a college algebra question.

00:35 Okay, so for 4 -8, you would go s of 8 minus s -of -4, divided by 8 -9 minus 4.

00:43 Because average velocity is the exact same thing as slope.

00:46 So s -of -8, i have 7, s -4 -7 over 8 minus 4.

00:52 So 0.

00:53 So the average velocity on 4 -8 is 0.

00:56 So it must be going forward and backwards, and they're canceling each other out.

01:02 Okay, now we're doing six to eight.

01:06 So that'd be wherever i have four, i'm going to put six now.

01:14 Okay, s of six is five.

01:17 So i get two over two, which is one.

01:20 So the answer to the first one was zero.

01:23 Answer to the second one is one.

01:26 Okay, now i'm doing eight to ten.

01:30 I don't want to scroll down farther because then i will have to keep going back and forth looking for things so i'm just going to erase what i had.

01:36 I guess you can watch it again.

01:38 All right, 8 to 10.

01:40 So s of 10 minus s of 8 over 10 minus 8.

01:46 As at 10 is 13.

01:49 S at 8 is 7 over 2.

01:52 That's 6 over 2, which is 3.

01:57 And then 8 to 12.

01:59 So this one will be s12 minus s8 over 12, minus 8.

02:13 So 23 minus 7 over 4.

02:17 23 minus 7 is 16.

02:21 So i get 4 for that one.

02:26 And in part b, you find the instantaneous velocity.