cdot-in-figure-2-27-you-walk-from-the-park-to-your-friends-house-then-back-to-your-house-what-are-yo

Question

$\cdot$ In Figure $2-27,$ you walk from the park to your friend's house,
then back to your house. What are your (a) distance traveled, and
(b) displacement?

Nathan Silvano · Accepted Answer

Okay, so in problem, too, we have to calculate the distance and the displacement. We&#x27;re still using the same figure off the problem one. But the difference now is that if the problem gives us and well, it was a centre problem, one distance is how much ground you cover while in motion. And displacement is the change in your position when you compare final position. If your initial position and is really important to know that your initial position is always zero off your coordinate system, okay, so you can have negative distance. You&#x27;re going to see this. You&#x27;re going to see that when you&#x27;re so the problem. Well, we have the distance between the park in the house. Judo 0.75 miles. Distance between library and park, zero point six miles. Distance between friends House and library off 0.35 Myers. Well, Robin gives us now a path that we go from the park through our friend&#x27;s house into the friend&#x27;s house, our house again. Well, the distance I just said. It&#x27;s just how much grown your cover. Then we have 0.6 from Proctor Library. Bless zero point 35 the library to friend&#x27;s house, plus Friend&#x27;s house to house. Ah, this is zero point five was zero point six of us, you know, point serving five. Myers. This is equal to two point 65 miles. Okay, then what is the displacement? Displace monies, the initial point and the final point? As you can see, we don&#x27;t go forward. We go backwards. So if you go backwards from our initial point or displacement is negative, Dan, what is the displacement? Place and meant is a co two minors. 0.75 Myers, the miner signed. It&#x27;s because we go backwards for any show points and known forwards or initial points. That&#x27;s problem one. Thanks for watching.

Nathan Silvano · Answer

so in problem one, we have a figure for reference, and the problem was to know the distance and the displacement when a given path. Well, to answer this question, we just need to understand what is the difference between distance and displacement. Distance is how much ground your object actually cover while, and emotion and displacement is the change in the final position compared the initial position. Well, let&#x27;s answer the problem. So this concepts will be more clear for you guys in this problem particular. Do we have a bark and a house separated war 0.7 five my house and a library in a park separated for zero point six Myers. Then the problem&#x27;s say&#x27;s that we go from the house, the library, then to the library, to the park. What is the distance that we cover? How much ground we actually cover? And his motion? Well, we go from the house for the park. 0.75 blows 0.6 The doctor The library plus 0.6 again. A library book that gives it that Give us Toto off one point 95 miles. This is our distance. This is how much ground recover in our motion. And what is the displacement? Well, as I said before, displacement is just a change to the final position. Danish A position. And this are Let&#x27;s see this placement. The displacement is just the distance between your house and a park. Udal point 75 That&#x27;s the true answers of the problem. Thank you for watching.

Surjit Tewari · Answer

question number one In the given graph, we can see that there is no change in that direction. Off the motion off the person and this is position. This is time be. Since there is no change in the direction of the person and the distance off the person from the original position is increasing continuously. So we can say that that distanced every day is equal to the magnitude off the displacement. From this graph, we can see that the original point is that 1 m and the final point is a six meat that so that the stands Trevor it will be 5 m big is equal to 5 m and this is also equal to the magnitude older. This place man, this 5 m answer.

Averell Hause · Answer

for part A. We confined the total ah distance traveled and this would be equaling 12 meters minus seven meters. This is equaling five meters. This would be your answer for part A for part B the magnitude of this placement. We can say that the magnitude of the displacement, um, would be equal to the shortest distance from origin to final position. So we have that the magnitude of displacement will call it the magnitude of X or weaken. Same magnitude of the attacks. This would equal again five meters and then report. See if it&#x27;s asking for the displacement. Well, the displacement now here is in the negative direction because particle be is traveling backwards. So the displacement would be equaling Delta X. And this is eagling, actually negative five meters. And so for a and B, the answer would be five meters for part C. Negative five meters. That is the end of the solution. Thank you for watching

Averell Hause · Answer

So here the distance for party the you can say the total distance traveled. This would be essentially nine meters minus three meters. This is equaling six meters and then we have five meters plus three meters another minus three meters. My apologies. This is equaling two meters and so the total distance traveled is equaling eight meters. This would be your answer for part a four part e The magnitude of the displacement which we could say that this would be equal to magnitude of Delta X would be equal in the shortest distance from the origin to the final position. So here we simply have four meters and then the displacement. For part C Delta X Aah! This would be equaling five meters minus nine meters. This is giving us negative four meters. So for party again, eight meters for part B, four meters for part C, negative four meters. That is the end of the solution. Thank you for watching

Cory Kuzinski · Answer

question 34. We&#x27;re trying to figure out our entire trip here. The first leg of the trip from our house to the amusement park we figured out in problem number 33. That was 269 miles. My work has shown on the next leg of our trip is after the amusement park. We go to the beach, so remember, the amusement park was at 100. You know, I&#x27;m gonna use in different colors here. Amusement park is at 100 to 50 and then we go to the beach. The fund A the beaches at 4 54 50 All right, So we used the distance formula to figure out the distance between those two points. So 4 50 minus 100 is 3 50 and we&#x27;ll square that 4 50 minus 2 50 is 200. And we&#x27;ll square that, um uh, use the calculator. Here, do 3 50 squared. Plus 2 50 square dot wants that&#x27;s 185,000. And then I think the square root of that, which is about, uh, 430 miles. All right. Now, we still need to figure out the distance between from the beach, which is that for 54 50 we go home, which is 00 So we need to figure out the destry in the beach in home. So we live 4 50 square plus 4 50 squared and then take the square to that. So I&#x27;m cheating a little bit for 50 squared plus 4 50 squared is 405,000. So we&#x27;re going to square root 405,000 and that distance is about 636 miles. So our entire trip IHS 269 miles to the amusement park, Um, plus 430 miles to the beach, plus 636 miles back home. All right, so to 69 plus for 30 who are 6 36 is a total distance of 1335 miles, while 1335 miles that is that does far

Problem

$\cdot$ The two tennis players shown in FIGURE $2…

Problem 2 Easy Difficulty

$\cdot$ In Figure $2-27,$ you walk from the park to your friend's house,
then back to your house. What are your (a) distance traveled, and
(b) displacement?

Answer

(a)
Distance Traveled: 2.65 $\mathrm{mi}$
Total Displacement: $-0.75 \mathrm{mi}$
(b)
The Answer is $[-0.75$ mi

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(a)Distance Traveled: 2.65 $\mathrm{mi}$Total Displacement: $-0.75 \mathrm{mi}$(b)The Answer is $[-0.75$ mi

Physics

Christina K.

Andy C.

Jared E.

Meghan M.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Christina K.

Andy C.

Jared E.

Meghan M.

(a)
Distance Traveled: 2.65 $\mathrm{mi}$
Total Displacement: $-0.75 \mathrm{mi}$
(b)
The Answer is $[-0.75$ mi

James S. Walker

Physics