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    Parts A and B of this following question make enough sense

    Name:  Q1 part C.JPG
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    But I'm lost when it comes to (C) and exactly what it's asking me to do. My current understanding of the question is that you could graph it like this:

    https://imgur.com/a/2gEOSQ0

    Which is that P and Q are velocities. with direction and magnitude. relative to a shared origin, 0. The question states that: At time t=0, the position vector of P is 400i meters and the pos. vec. of Q is 800j metres. My understanding is that at t=0 and with a shared origin, both cars should be at the same point. And also, what is 'due west'? exactly west of a point on P or just somewhere in the vicinity? And when exactly does Q ever actually reach west of P if they're both moving east?

    Thanks for any help
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    due west is when the vector QP is parallel to the x axis (ie j component = 0)
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    (Original post by Dggj_19)
    Hi all, I've just started going through M1 for the first time in preparation for my resits next June. Parts A and B of this following question make enough sense

    But I'm lost when it comes to (C) and exactly what it's asking me to do. My current understanding of the question is that you could graph it like this:

    https://imgur.com/a/2gEOSQ0

    Which is that P and Q are velocities. with direction and magnitude. relative to a shared origin, 0. The question states that: At time t=0, the position vector of P is 400i meters and the pos. vec. of Q is 800j metres. My understanding is that at t=0 and with a shared origin, both cars should be at the same point. And also, what is 'due west'? exactly west of a point on P or just somewhere in the vicinity? And when exactly does Q ever actually reach west of P if they're both moving east?

    Thanks for any help
    I'm not sure why you reason that both cars should be at the same point at t=0?? One car is 400 metres up the graph from the origin, the other is 800 metres to the right of the origin when t=0.


    Anyway, when Q is due west of P, then you need to look at P and find the time when Q is to the left of it. And not just 'in the vicinity' as you say, it must be straight left of it.

    With a simple sketch of the situation when Q is due went of P, you should notice that they share the exact same \mathbf{j} component. Simple enough intuitively because when you travel from P to Q, going westward, you do not need to go up or down the graph at all. Hence that vertical component remains unchanged.

    You can use this fact along with answers of part (b) to deduce an equation in t to solve. Notice that \mathbf{p} is the position vector where P is at time t, and likewise \mathbf{q} is the same for Q.
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    (Original post by Arctic Kitten)
    due west is when the vector QP is parallel to the x axis (ie j component = 0)
    Vectors have magnitude and direction, though, so if a combination of Q and P happened to be exactly parallel to the X axis (I'm guessing this question is structured so that's intentionally the case), would it not point east? I'm also a little confused as to why this requires you to combine/equate them when the question implies they're still separate/independent.
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    The vector QP is vector P minus vector Q. Their sign (the sign of the i component) must be positive.
    What vector QP mean is the relative position of P from Q, and you want that position to just be east (Q is west of P so P is east of Q) ie consists of only positive i component.
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    (Original post by RDKGames)
    I'm not sure why you reason that both cars should be at the same point at t=0?? One car is 400 metres up the graph from the origin, the other is 800 metres to the right of the origin when t=0.


    Anyway, when Q is due west of P, then you need to look at P and find the time when Q is to the left of it. And not just 'in the vicinity' as you say, it must be straight left of it.

    With a simple sketch of the situation when Q is due went of P, you should notice that they share the exact same \mathbf{j} component. Simple enough intuitively because when you travel from P to Q, going westward, you do not need to go up or down the graph at all. Hence that vertical component remains unchanged.

    You can use this fact along with answers of part (b) to deduce an equation in t to solve. Notice that \mathbf{p} is the position vector where P is at time t, and likewise \mathbf{q} is the same for Q.
    Thanks for such an indepth response. However, you've mentioned several things that I'm still having trouble rationalising. First of all, you mention that I need to look at P and find a time when Q is left of it - that is, exactly horizontal to the point on Q. Looking at what I've drawn, there's no point on P from the origin 0 to infinity, where a point on Q is directly horizontally left of the point on P, no matter the initial position of either car along either vector. This is what causes me to not understand your second point, which involves a sketching a situation where Q is due west of P. I understand that you can express p and q as 400i + t(15i + 20j) and 800j+ t(20i -5j) respectively, but after that, I'm still lost.
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    (Original post by Dggj_19)
    Thanks for such an indepth response. However, you've mentioned several things that I'm still having trouble rationalising. First of all, you mention that I need to look at P and find a time when Q is left of it - that is, exactly horizontal to the point on Q. Looking at what I've drawn, there's no point on P from the origin 0 to infinity, where a point on Q is directly horizontally left of the point on P, no matter the initial position of either car along either vector. This is what causes me to not understand your second point, which involves a sketching a situation where Q is due west of P. I understand that you can express p and q as 400i + t(15i + 20j) and 800j+ t(20i -5j) respectively, but after that, I'm still lost.
    I fear you might've drawn the wrong thing which is what confuses you. Here is a link to help you see what is really going on.

    https://www.desmos.com/calculator/ylwetsnyjg

    As you play with the slider for T, which is the time, the particles will move to their locations at that time. Moving the time to about 32 seconds you will notice that indeed Q is to the left of P.
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    (Original post by RDKGames)
    I fear you might've drawn the wrong thing which is what confuses you. Here is a link to help you see what is really going on.

    https://www.desmos.com/calculator/ylwetsnyjg

    As you play with the slider for T, which is the time, the particles will move to their locations at that time. Moving the time to about 32 seconds you will notice that indeed Q is to the left of P.
    I think this makes me understand my error. 800i and 400j are the initial positions of the vectors but the magnitude and direction of the vectors haven't changed, meaning that there is actually a point wherein Q can be directly horizontally left of P. Given this, what exactly is the question asking when it says 'find the position vector of Q when Q is west of P'? Is it asking to find the value of t for which Q is west of P? And do you do this by equating the Y (j) components and finding t?
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    (Original post by Dggj_19)
    I think this makes me understand my error. 800i and 400j are the initial positions of the vectors but the magnitude and direction of the vectors haven't changed, meaning that there is actually a point wherin Q can be directly horizontally left of P. Given this, what exactly is the question asking when it says 'find the position vector of Q when Q is west of P'?
    I'm not sure I can break down the question any more than it already is?? Just, whenever Q is west of P, what are its coordinates, i.e. position vector ? This happens only once in their motion, as you can observe from the link, so when it DOES happen, what are Q's coordinates?

    Of course, to do that, as I mentioned before, you need to determine the time WHEN this occurs. Then sub it into the vector \mathbf{q}
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    (Original post by RDKGames)
    I'm not sure I can break down the question any more than it already is?? Just, whenever Q is west of P, what are its coordinates, i.e. position vector ? This happens only once in their motion, as you can observe from the link, so when it DOES happen, what are Q's coordinates?

    Of course, to do that, as I mentioned before, you need to determine the time WHEN this occurs. Then sub it into the vector \mathbf{q}
    I edited my last post with a question to clarify if I understood your explanation. I tried it out and finally I understand my error and got the correct answer. Thank you so much for your help and again, I'm sorry it took so long - I'm a private candidate teaching myself M1/M2 and this is my second day of studying it, so I'm still very new to Mechanics as a whole
 
 
 
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