Vectors.

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ganeshie it's kind of the same thing we just did except for without angles given.

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draw both the velocity vectors break them into component form and simply add them up
|dw:1433310949412:dw|
wait, I forgot the component form.
|dw:1433311043791:dw|
so 19
Nope, you need to add the respective components
x components : -5+0 = -5 y components : 0+24 = 24 so the resultant velocity vector with reference to water is -5i+24j
haha right. sorry. So now how do we change that to decimal numbers? (note how the first question wants to be answered)
simply take magnitude of the resultant velocity vector for speed
speed = \(\|-5i+24j\|=\sqrt{(-5)^2+24^2}=?\)
\[\sqrt{601}\] =24.52
looks good
fabulous. part next? :)
work the angle for any vector \(xi+yj\) the angle with the positive x axis is given by \(\tan^{-1}(y/x)\)
so since the five is negative..do i carry the negative with it?
78.23
try 11.77 for angle
yeah that worked :P
how did you get there?
you got angle = \(-78.23\) right, it is negative
in which quadrant does the angle \(-78.23\) lies in ?
4
yes, but the resultant vector lies in 2nd quadrant : |dw:1433312298520:dw|
so you need to add 180 to the angle given by your calculator : \[-78.23+180=?\]
101.77
Yes thats the angle from positive x axis : |dw:1433312428785:dw|
but the question is asking for the angle from N : |dw:1433312468055:dw|
oh oh i see.
so that - 90 which comes out to 11.77
thats it! with \(\tan^{-1}\) function you always need to check if 180 degrees needs to be added based on the quadrant of the actual point
so if it's negative then always yes add 180 yeah? :P
not quite
tell me this in which quadrant does the point (-1, 1) lie in ?
2
so would you trust me if i say the angle is -45 with positive x axis ?
try working the angles using arctan(y/x) formula for below points : (1, 1) (-1, 1) (1, -1) (-1, -1)
45 -45 -45 45
do they make sense ?
look at the last point and last angle : (-1, -1) 45 how can a point in III rd quadrant make an angle 45 ?
yeah.
these are good questions to think about :o
it doesn't make sense, so you need to add 180 to 45 : 45+180 = 225 so the angle made by vector (-1, -1) is 225 and not 45 ok
\(\tan^{-1}(y/x)\) formula gives you correct angle when the point is in I or IV quadrants
for III and II add 180?
When th point lies in II or III quadrants, you need to add 180 degrees to the answer given by your calculator
In our case, the point (-5, 25) is in II quadrant so we added 180 degrees to the answer given by calculator
this is one of the things that trips all students when they begin to learn inverse trig functions
yeah man, i'm kind of trippin now haha but I think I get it.
watch this video if u have time https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/inverse_trig_functions/v/inverse-trig-functions-arctan
not now, when ever you have time and feel like... it explains arctan function very nicely
Thanks :)

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