A cat walks in a straight line, which we shall call the x-axis, w...

Question

A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat’s motion and construct a graph of the feline’s velocity as a function of time (Fig. E2.30). (c) What distance does the cat move during the first 4.5 s? From t = 0 to t = 7.5 s? (Image)

Accepted Answer

Hey everyone. So today we're dealing with some velocity versus time graphs. And we sort of have to analyze them a little bit. So we're being asked to determine the distance covered during the first three seconds and the distance covered from time two seconds, two times six seconds during a jog that is motioned. Or that is measured where the motion is parallel to the X axis and to the right in the positive X direction, the positive X direction. So let's go ahead and just jump right into it. So the first thing we need to note is that the distance covered is equal to the magnitude of the area under a velocity time graph. Right? So distance is equal to area under V. T. Graph. So what does this mean? Exactly? Well, let's sort of break this up if and I'll use blue for the first section until three seconds. Right here. The area under the graph would be between the graph and between the X axis where it intersects right? And it's negative. But the area that we're looking at is this area. Oops, I didn't try that. Right. This one tips, it's this area right here. And as you can see it's a little trapezoid all all in here. So this is the area that if we calculate this area will get the distance traveled during that time period. So let's do that. So the area of a trapezoid can be calculated as 1/2 A plus B into the height where a and B. Are the two bases of the trapezoid. Now for it to look at this, the basis would be A. And B. And the height would be this the the distance on the X. axis from 0 to 3. So let's write that out. It would be 1/2 in two. Well from 0 to -1 Is well -1. From 0 to -4 is -4. So a. Would be -4, negative four plus negative one in 23, which one added up, gives us an answer. Well once you multiply it my bad And then once we simplify everything, we get a final answer of negative 7.5 m. And since we're calculating distance, The distance will simply be the absolute value. So distance will be the absolute value of negative 7.5 m. Which is simply 7.5 m. Now we can do a similar thing from points or from time to until time six. Right. And here we have two areas that are now in the shape of two triangles. So let's look at this in A. I'm gonna scroll down so we have a little more space and let's use green. So from time to until time six it can be segmented into two separate triangles. And I'm going to erase my blue trapezoid here. Let's draw it in green we have two triangles. We have one this triangle right here and this triangle right here it's a recall for triangles area of a. Of a triangle is equal to one half base times height, one half B. H. So if we're talking about the total area well that would be one. F. B. H. Plus one half B. H. Excuse me. So for And we'll say this is area one And this is area two. Let's label these as well too. Area one. Now for area one it's going to be the change in velocity over the change in time right? Or the change in velocity into the change in time, I should say. So the change in velocity Goes from -2-0. So that will be negative two. So it'll be one half into negative two while the height is from ah Excuse me, I did that backwards from zero to negative two here for the velocity while the time goes from two 24 which means it'll be negative too into two says a one and that will equal -2.0 m. So this is for a one for a two. We use something very similar. The change in Velocity goes from 0 to 2 so it'll be one half into two multiplied by the change in time goes from 4 to 6 which is again two seconds in the positive X direction, That'll equal two m. Oops that went off the screen. 2.0 m. However, recall that when we add up the areas since we're doing displacement right? The distance traveled, we're not talking about if he's gone forward or back. So it's not displacement. It is the total distance traveled. We have to take the absolute value of both and then we add them together so that becomes is equal to a one plus a two, which equals the absolute value of negative two point oh Plus that's a value of 2.0, which equals four 0. m. So the correct answer choice in this situation would be answer choice. A. The distance traveled from time 0-3 seconds with 7.5 m And the distance traveled from time to to six seconds was four m. Alright, the leader. four m. I hope this helps. And I look forward to seeing you all in the next one.

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