the-golfer-in-figure-2-29-sinks-the-ball-in-two-putts-as-shown-what-are-a-the-distance-traveled-by-t

Name: the golfer in figure 2 29 sinks the ball in two putts as shown what are a the distance traveled by t
Uploaded: 2020-02-08
Duration: 2 min 17 s
Channel: Numerade
Description: The golfer in FIGURE $2-29$ sinks the ball in two putts, as shown. What are (a) the distance traveled by the ball, and (b) the displacement of the ball?

Question

The golfer in FIGURE $2-29$ sinks the ball in two putts, as shown. What are (a) the distance traveled by the ball, and (b) the displacement of the ball?

Nathan Silvano · Accepted Answer

So and this problem, we have a go for who thinks about in two boats. Let&#x27;s see. We have this distance of 10 meters and another distance off, too 0.5 meters. So their first boot, the ball&#x27;s goes from here until here and the last one. The balls goes from here, open to the sink. So the problem wants us to find out what was the distance that the ball travel and the displacement. Well, let&#x27;s see things who, since we already know distance, is just sum up the two errors. So we just need to some ups. That&#x27;s I just need to some 10 plus 2.5, plus the second arrow off another 2.5 meters. So the total is just a total distance that above travels is just 15 meters. And what was the displacement? Displacement, we know is the Delta X. Delta X is just the final position of the ball, minus their initial position off the ball. The initial position off the ball was the orange of the movement. The orange of the movement was right here, and zero so x zero is just zero, and the final position of the ball is just right here at 10 meters of distance from the origin. So I was just going to be 10 minor zero. So the displacement off the ball is just Dan meters and that&#x27;s the final answer for this problem.

Nathan Silvano · Answer

so in problem. Three. We want to know the distance and the displacement off the two tennis players when they walk to one match to congratulate one another. Well, this problem is basically we have the net. The distance between the nets and the tennis player B is two meters. The distance between tennis plea or a is five meters for the net. Okay. So again, distance is how much ground you cover while you are in motion. And displacement is just the difference between your position, your final position in your initial position. Okay, so the first item we want to know about a perspective off tennis player, eh? His distance is five meters, because is just how much ground it covers. Why are all contained in that? And his displacement is also five meters because is final position. It&#x27;s five in his initial position. Zero Okay, well, second night and we want to know about a perspective off tennis player B. Well, he&#x27;s distance is true meters because he&#x27;s just over two meters of ground and his displacement is also meters because his final position is too. And his initial position is zero. Okay, Best define our answer. Thank you for watching

Nathan Silvano · Answer

So in problem 24. The first I think we just have to tell which off the segments has festered. The greatest speed. Wow. We know that the greatest speed is associated with the fastest growth. And we can see this segment which has the festive growth. Is this segment be so? The answer for the first item is B because he has the faster growth. Okay, so do you other items the itin be see and d We have to calculate the extra value off the velocities and compare with our physical knowledge of the system. So we know that average velocity is defined as the difference in position, divided by the difference in time. So we can see that velocity off the first segment. It&#x27;s going to be the difference. A position which is the girl from position Tree from position one. It was just one minus tree divided by how much seconds? Two seconds. So this is it goes to minors. One meters 1st 2nd velocity off this segment be is we go from position one back to position three. So we have three. My no swan divided by one second. This is just two meters per second and the velocity of the segment C. We go from position to for position. No, Sorry. We go from position three to position too. So girlfriend position three from position to in two seconds. So this is miners. One divided by two are minor, 0.5 meters per seconds. Well, as you can see, the fastest speed is in B, which also has the fastest growth. That&#x27;s the final answer. Thank you for watching.

Patrick Connors · Answer

So in this problem, we have to soccer players that are standing 48 meters apart and they both start from rest before running towards each other. And we&#x27;re looking for how long it takes them before they collide as well as how far the first player is running because we&#x27;re analyzing the motion of two different players were goingto have two different lists of the Kinnah Matic variables. Here, here is player one who is running to the right and player to who is running to the left. Now the information we know here it&#x27;s a little bit limited. Player one is starting at zero meters per second because he&#x27;s starting from rest of the same as player, too. We know the acceleration of player one is 0.5 meters per second and for player to its 0.3 meters per second squared now because player too, is running to the left. That means it&#x27;s going to be negative 0.2 meters per second squared. Okay, then we also know the time of their collision is going to be the same thing and the distance between them in total, is 48 meters. Now what this means is player, one is going to be running forwards and player too is going to be running backwards. This total distance between them is 48 meters. So player one is going to run forward some distance, which I&#x27;m going to call ACS now. Player too is going to run the other part of that distance, which is gonna be 48 minus acts because they&#x27;re running However much player one doesn&#x27;t run. Um, the tricky piece here is that since player too is running to the left, it&#x27;s going to be a negative value. So it actually ends up being that player, too is running X minus 48 meters. So that way it&#x27;s definitely going to be a negative number. And it&#x27;s the magnitude that would add both those up to be 48 meters guys. So that is everything we know right now. Um, I kind of know their values and have set up X related to each other. Now for both of these, we want to know the time and the distance between them, so we don&#x27;t know their final velocity, and we also don&#x27;t need to find their final velocity for either one of these So when I&#x27;m looking at which equation I want to use, I&#x27;m gonna be using the one that does not have Final velocity, which is the last one on the list that X equals V not t plus 1/2 A T squared. So for both player one and player, too, we&#x27;re starting with X equals V, not tea plus 1/2 A T squared. So let&#x27;s d&#x27;oh player one. When I plug in my numbers here we have X equals V, not t, which since the initial speed is zero, this whole term just goes to zero plus 1/2 of the acceleration. So 1/2 of 0.5 gives us 0.25 multiplied by the time squared. So writing this down in nice simplified fashion player one is X equals 0.25 t squared Got So that&#x27;s everything we can do for player one using that equation? No, no said that aside here, let&#x27;s do the same work. But looking at what player to condo&#x27;s so player, too. The distance player two runs is not acts. It is X minus 48 equals V, not t will. He also starts with a velocity of zero plus half of his acceleration. Um, so that&#x27;s going to be negative. 0.15 multiplied by the time squared. So then I have X minus 48 equals negative 480.15 I want to get X by itself. So we have X equals 48 minus 0.15 times T squared. Okay, Now, looking at this, I have solved both of these equations in terms of acts. And since acts is the same distance for both of them, I can take my player one equation and my player to equation and use the substitution principle to set them equal to each other. So I&#x27;m going to do that and continue my work here. Let me just move this out of the way. Okay, there we go. So when I put this together, we get 0.25 t squared is equal to 48 minus 0.15 t squared. So when a group my like terms. So I add 0.15 t squared to both time sides and we get 0.40 t squared equals 48 divided by 480.40 and we get t squared equals 1 20 And then take the square root of that to get our time of 11 seconds. So it takes 11 seconds for these two players to bump into each other. Then taking that 11 seconds, I&#x27;m gonna plug it into our player one equation that X equals 10.25 t squared and use that to figure out the distance between them where the distance player one runs when they collide. So we have 0.25 times 11 squared and we end up with a distance between them of 30.25 meters, meaning player one runs 30.25 meters to the right and collides with player to after 11 seconds.

Averell Hause · Answer

So for party, we have the average were simply using the figure 2.8 that they give you in the, uh in the textbook. And essentially we&#x27;re saying that the were trying to find average velocity. This is equal to the change in this plate or the displacement in the ex direction with respect of time. And so we can say for part A Ah, from zero seconds to, um one second, we essentially have the average velocity equaling four meters minus zero meters, divided by one second, minus zero seconds. And so the average velocity is giving us four meters per second. This would be your answer for party four Part B. We&#x27;re trying to say zero. We&#x27;re trying to find the average velocity from zero seconds ah, to four seconds. And so we can say them that the average velocity is gonna be equaling negative two meters, minus zero meters divided by. And then this would be, of course, four seconds minus zero seconds. This is equaling negative 00.5 meters per second. So this would be your answer for part B four reports. See, we want to find, uh, the time from the rather the average velocity from one second to five seconds. The average velocity here would be equaling zero meters minus four meters, divided by five seconds, minus one second. And this is giving us exactly negative one meters per second. This would be our answer for part C and finally four part D. Ah, We have zero seconds to five seconds. So our average velocity is gonna be equaling zero meters minus zero meters again. Here are displacement is zero. So here, no matter what our time is, we&#x27;re going to get an average velocity of zero meters per second. This would be our answer for part D. That is the end of the solution. Thank you for watching.

Sarah Wharton · Answer

essentially have a triangle. We know that this is 20.3 feet. We know this is 19.4 feet and we know the angle here is 56 degrees and ah, we want to know what this angle angle Alfa is equal to. Wilful will first need to find what angle beta is by using love signs. So, um, it&#x27;s gonna move this out of the way area. All right, So we&#x27;ll dio sign of beta divided by 20.3 is equal a sign of 56. Divided by 19.4 cross multiply to get 19.4 times. Sign of beta is equal to 20.3 times Sign of 56 by both sides by 19.4 to get sign of beta is equal. Teoh 20.3 times. Sign of 56 divided by 19.4. Then take the inverse sign of both sides To get beta is equal to 60.169 degrees. And then, uh, if you subtract, um, you&#x27;ll get Alfa is equal to 180 minus 56 minus 60.169 degrees. So Alfa is equal. Teoh, 63.8 31 degrees. That&#x27;s it

Problem

$\cdot$ A jogger runs on the track shown in FlouR…

Problem 4 Easy Difficulty

The golfer in FIGURE $2-29$ sinks the ball in two putts, as shown. What are (a) the distance traveled by the ball, and (b) the displacement of the ball?

Answer

(a)
a. 15 $\mathrm{m}$
b. 10 $\mathrm{m}$
(b)
The Answer is $[10 \mathrm{m}]$

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(a)a. 15 $\mathrm{m}$b. 10 $\mathrm{m}$(b)The Answer is $[10 \mathrm{m}]$

Physics

Christina K.

Andy C.

Liev B.

Marshall S.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Christina K.

Andy C.

Liev B.

Marshall S.

(a)
a. 15 $\mathrm{m}$
b. 10 $\mathrm{m}$
(b)
The Answer is $[10 \mathrm{m}]$

James S. Walker

Physics