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Problem 11 Medium Difficulty

(a) A particle starts by moving to the right along a horizontal line; the graph of its position function is shown in the figure. When is the particle moving to the right? Moving to the left? Standing still?

(b) Draw a graph of the velocity function.

Answer

(a) The particle is moving to the right when $s$ is increasing; that is, on the intervals (0,1) and $(4,6) .$ The particle is moving to the left when $s$ is decreasing; that is, on the interval $(2,3) .$ The particle is standing still when $s$ is constant; that is, on the intervals (1,2) and (3,4).
(b) Velocity function graph is slope of the position vs time graph. The velocity is 3m/s on (0,1); 0 m/s on (1,2) and (3,4); -2 m/s on (2,3); 1m/s on (4,6)

Discussion

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CA

Catherine A.

October 26, 2020

Daniel J., thanks this was super helpful.

CA

Catherine A.

October 26, 2020

SA

Sharieleen A.

October 26, 2020

This will help alot with my midterm

Video Transcript

This is problem number eleven of the Stuart Calculus eighth edition. Section two point seven party. A particle starts by moving to the right along the horizontal line. The graph of its position function is shown in the figure below. When is the particle moving to the right? Moving to the left. Standing still. So if we look at the blood here, um, that is similar to deploy it in the book, it says that the particles start moving to the right if it starts moving to the right, Then we assume that as it moves up this position function iss, it is willing to the right. Therefore, if it's moving backwards town, it is moving to the left. And if it's not showing anymore mineral, that's when it's standing still. So we can see here that as the function, no moves initially moves to the right, it keeps moving to the right until he reaches one second at one second and stop one to the right. Um, where it now changes from one for moving to the right. Two standing still. So it goes from moving to the right for the first second two standing still for the next second from one to two for the next second. We're moving down in this graph. Down or backward on this function graph means that we're moving to the left opposite of where how we started. So we're moving to the left of term time he goes to technical three and then we stand still once again from time three to temp for because we do not change our position. Finally, from ten Port Attempt six, we are moving up and positively in the positive ass direction, meaning that we're winning in the initial direction that we started with, which is right. So we're willing to write again from the time he goes for two technicals six. Yeah, For part B, we're drawing a graph of the velocity function. Velocity is related to position has its slope. We're gonna use the slopes of the graphic for position to craft the velocity function. And since our function in in the position graph can be thought of as separate, separate linear function. So we have one line here, one straight line and another straight line that another straight line. Each of these straight lines has a constant slope so underground for the blasting function will have constant values for each of these time periods. Okay, so for the first, second and the first part of the craft we have a slope of is this is a straight line and I had the slope of three over one as having to stop of zero. So these are the flat parts of the graph. So when extending still, we go from I've lost three immediately. Timing goes one down to Rosset of zero from zero from one to two seconds afterward. From two to three seconds. We have ah, this negative slope and calculated as native to over one rise over run. That's negative too, to slip up negative too. And the negative velocity for physical purposes means that you're moving in the opposite direction. So this court, this is consistent with moving left in this problem and where we have a soup of negative too for this learning up until ten. Because three seconds where we have another zero slope, we are standing still again. No velocity Finally from chemicals for second Systemic yl six seconds. We have this new line that at the start of one to this slope of one is equal to the velocity of one from four to six seconds and if they constant slope, so it's a constant velocity from four to six seconds. Therefore, the function here to the right that we just drew is the velocity function for this given position function.