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Question

An object travels along the vector $d_{1}=4 \mathrm{m} \hat{\mathbf{i}}+5 \mathrm{m} \hat{\mathbf{j}}$ and then along the vector $d_{2}=2 \mathrm{m} \hat{\mathbf{i}}-3 \mathrm{m} \hat{\mathbf{\jmath}}$ How far is the 5 $\mathrm{j}$ object from where it started?
(A) 6.3 $\mathrm{m}$
(B) 8 $\mathrm{m}$
(C) 10 $\mathrm{m}$
(D) 14 $\mathrm{m}$

Sachin Rao · Accepted Answer

problem we have given an object travels along the Vector demon, which is four metre in the cab direction plus five m in the Jacob direction. And victor D two is given two m in ICAP direction and minus three m in the Jacob direction. So we have to find the distance from the object where it is started. So we know that total displacement will be called to do even plus the two. So this is a call to four plus +26 icap. Let&#x27;s fight minus three. So this is blue Jacob. No. Yeah. We have to find the magnitude into divan. Plus in D two is a call to on the road six squared plus two square. So this is a call to around 6.3 mi that No, If we check our options option ocean able, we correct it is known that tank

Ze-Han Lee · Answer

So in this problem, we have two factors. So the one is full. I plus five j fray and a D two is too. I minus three j. So we have 10 that a total displacement D is d one pastilles too, which is Ah, six I plus two j. And the magnitude of this displacement is ah, square root. Six quid plans to squid, and this gives you 613 All right.

Zachary Warner · Answer

So our question says that we have, ah, two displacement vectors, one of the magnitude of three meters and one of a magnitude of four meters. And it wants us to figure out how we need to combine these two displacement vectors to get a vector of a magnitude of a seven meters. Be one meter and C five meters. Okay, well, in order to get seven meters from two vectors that air three meters and four meters, we can see that. So this is part eh? That&#x27;s seven meters is equal to three meters plus four meters. So therefore, these vectors must be well. Just type it out here, parallel in the same direction, become box. It is their solution or part, eh? Part B wants us to figure out what we need to do to get one meter from these two. Well, clearly, one meter is equal to four meters minus three meters. So therefore they have to be parallel, but in the opposite direction. Lincoln box it in as our solution, or B. Okay. Okay. Now, for part C, we&#x27;re supposed to get one that is equal to five meters. Okay, But in order to get five meters. We can realize that this is equal to the square root of three meters squared plus four meters squared. Okay, so the square root of three meters squared plus four meters squared is equal to five meters, so therefore they must be perpendicular to one another. So I just kind of show you what I mean. Here you have to be like that. So who? It&#x27;s not very good. So we&#x27;ll say this is four. This is three Miss would be five. Therefore there perpendicular. All right box and insert solution.

Andy Chen · Answer

were given vectors A and B in terms of their ex unit vector and one unit vectors, and the Roman wants us to find the magnitude and directions of these two vectors here. Let&#x27;s start with a which I labeled them red and we&#x27;ll move on to be which I believe will in green. For a the magnitude, we can simply use the Pythagorean theorem to find the high partners H A of this right triangle, So H a is equal to the square room of 22 meters square, close to 14 meters negative 14 meters square. That gives us gave attitude about 26.1 meters to find the angle, which is this angle here, theta of a. We can use the inverse tangent function of why a over X a well, why is negative? 14 meters and exit is 22 meters and we get an angle of negative 32.5 degrees. Similarly, for B, we used without hearing serum on this right triangle here, HB will be able to square room of 2.5 meters squared plus 13 meters squared to give us a magnitude of 13.2 meters and to find the direction we apply the cinema process of inverse tangent. Why be over X B and get an angle of 79.1 degrees? Now the problem wants us to find the magnitude and direction of a plus B. Here we start with a and talk on getting the letter b Such a the resulting vector. We will label as C and is given by the blue factor. Now we can also write see in terms of the ex hat and why have so if her ex hat We will take 22 plus 2.5. So we get 24.5 meters in the X direction and native 14 plus 13 which is negative one meter in the one direction to find the high partners of this factor to get its magnitude. Well, you get used to the greens there. The manager will be the square of 24.5 meters squared plus needed one meter square, and that still gives us the 24.5 meters to find the angle, Which is this one right here? We have inverse changing of why see over exceed, and we get an angle of about native 2.3 degrees

Ze-Han Lee · Answer

so we won&#x27;t find out the total displacement for this ah movement so that it will be simply the area enclosed by the curve. Okay, so we can just to divide them into different shapes. Like what I did is that from a to B the areas three times seven over too. And from me to see the areas seven times two and the sea duty is ah, seven plus three plus seven times well or two. And the D to e is ah, three times three over two and e to f biggest from eat after the velocities. Negative ray. So she used mine is here minus two times two or two. So this gives you a 32 meters Vitamin E. Yeah.

Susanna Weber · Answer

here we have Victor A, which is five by minus two. J Director. Be with components minus three. I minus five J and we want to find a vector D, which is equal to a my speak. First, we want to just write our vector in component form. So we&#x27;re going to subtract the different components of our victories. We have five, I minus negative three. I just gonna be a I. We have minus two J minus negative five J, which is gonna give us three j since two negatives make the positive. Next, we&#x27;re being asked to draw out our vectors. We know that since a has to next has a positive I in negative Jake opponent. It&#x27;s going to be pointing here positive X direction and negative y direction In the fourth quadrant. B has a negative I component and a negative. J Components is reporting here into the third quadrant with X and y both decreasing and we can see by vector addition. Rules are visual vector addition rules How d is, in fact, a minus bi. So we have negative B is in reporting, of course, in the negative direction of our original vector. So pointing up here. And as you know, we can shift vectors in space as long as we&#x27;re not rotating them. So you could also draw negative, be here. Just run with these black dots, and then d must be after we add from to detail it would look something like this. Which makes sense, since both are I and J component of g r. Positive. So this is, in fact, about the direction that it would point it bright. See, Let&#x27;s first find the magnitude. So we&#x27;re gonna add our components, square them and then square root them. So we&#x27;re gonna eight squared plus three squared, which give us a magnitude of the square of 73. To find data as usual, we&#x27;re gonna take the inverse tangent over y component over x component. So we&#x27;re going to get 3/8, and that&#x27;s going to give us a result of about 20.57 degrees where direct

Problem

If $\mathbf{A}=\nearrow$ and $\mathbf{B}=\nearrow…

Problem 7 Easy Difficulty

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