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(II) You are driving home from school steadily at 95 $\mathrm{km} / \mathrm{h}$for 130 $\mathrm{km} .$ It then begins to rain and you slow to 65 $\mathrm{km} / \mathrm{h}$ .You arrive home after driving 3 hours and 20 minutes.(a) How far is your hometown from school? (b) What wasyour average speed?

a) $2.6 \times 10^{2} \mathrm{km}$b) $77 \mathrm{km} / \mathrm{h}$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Winnipeg

Lectures

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this's Chapter two problem six. So in part a of this problem, we're looking to find the distance from school to your hometown, and we're given some information about the trip forgiven the total time the trip takes. And then we know that the trip has two distinct legs. So each leg of the trip is some travelling at constant speed, which is the sort of the simplest type of motion that we can work with in this chapter. But the two speeds are different, so we can't just trip treat the entire trip as emotion, a constant speed in the first leg of the trip. We're going at 95 kilometers per hour, so we know our speed. We know that the distance taken is 130 kilometers, but we don't know how much time that way takes the second leg. We know you're going at 65 kilometers per hour, but we don't know how much time that takes, and we don't know what it's distances we do know. However, the total time that both legs took together. So that means if we could find the time for one leg, we could find the time for the other leg. So let's try doing that. Looking at the first leg first, we know the speed and we know the distance. So we know from our good old equation in general quips, Let's do this in black. Actually, this's just a general equation not pertaining to any of the legs of the trip. That average velocity is displacement, divided by time, Time elapsed. So in this case, for the first leg, we know Delta acts the distance and we know Delta V and are trying to solve for Delta T. We can rearrange this to say What else did he equals? Delta X. That's our 130 kilometres divided by average speed. That's the 95 kilometers per hour and so we can plug those in and you'll notice thie kilometers air going to cancel the hours or in the denominator of the denominator, which is the same as being in the numerator. So you end up with one point 37 ours. All right, that's the time for the first segment. So if you want to find the time for the second segment, we can subtract it from the total time. First, you'll notice that our time there not quite in the same units. We've got our time for the first segment and read in a decimal number of hours and in green we've got it in hours and minutes. So it's convert the hours and minutes two decimal of ours. There's 60 minutes in an hour to 20 minutes is 1/3 of an hour or 0.333 repeating hours, so you can write that as 3.33 hours, and that means the time for the blue trip is going to be 3.33 hours. It took time for both or did time with Blue Lake going to be the time for both legs. Together, this 3.33 hours minus just the time for the first leg. The 1.37 hours and that will get you 1.96 hours, or about one hour, 58 minutes so you can put that up here all right, great. So that's useful when we know the speed and the time for each leg. But we're not. What we're looking for ultimately is thie distance, so we know we already know the distance for the first leg. That's our 1 30 kilometres. So now we just gotta find the distance traveled in the second leg and we'll add that to the 1 30 kilometers. That's a total distance. So here notice. Look back at our equation, relating average velocity and displacement and time. Now it's the displacement that we're trying to solve for, so we can get that by Delta X rearranged equation to Delta X is average velocity, the bar times, Delta T and we plug in the 65 kilometers per hour to our average velocity within 1.96 hours to our delta T, and we find that our Delta X is going to be one 27.5 kilometers. So for a total distance, we've got to take the 1 30 Add it to the 1 27.5 and you get 2 60 kilometers. And here I'm rounding some only human too significant figures because you might also write. This is 2.6 times that of the two kilometers because they only had two significant figures in our input. But we've been keeping Maura's intermediate steps so we don't end up getting actual errors and our answer do surrounding. But the answer you should report to. The best of your knowledge is 260 kilometers for the total distance, and that's per day. Now, in part B, it wants What was your average speed? So now we're looking for average speed over the entire trip. Noticed that you don't get that by averaging the two speeds that you were travelling at. Because if you think about it, your let's say you were travelling at 95 kilometers per hour for a bit and then 65 kilometers per hour for a bit. Your average speed is going to penned on for how long you were travelling 95 kilometers per hour and for how long you were travelling at 65 kilometers per hour. So you can't just average the two speeds because you had a lady of speed. Instead, what we can do is go back again to this equation, says average speed or an average velocity, really, of which the average speed of the magnitude is going to be for any given chunk of travel. Average speed is thie displaced, the distance traveled, divided by the time elapsed. So in this case now we've had to calculate the distance traveled and we were told the time elapsed so we could just get that Our overall average speed is the quotient of those two things. And now So we're taking both legs of the journey together as one piece of travel and looking for the average speed. So 2 60 kilometers divided by this 3.33 hours. Actually sorry, we really should be doing here. We rounded to 2 60 but should put in two more. The more accurate numerical value that we had from adding 130 kilometers in 127.5 kilometers again to avoid round here's. But now you divide this. Now you report your answer to two significant figures. So you've got 77 and units air just kilometers per hour, and it's quickly check. We didn't. We knew that we couldn't just average 95 65 to get our answer kilometers per hour. But you know that the average speed should be somewhere between those two speeds, and it is 75 7 kilometers per hours. He was perfectly reasonable, and that's your answer

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