$v_{tot_x} = 4.41 \textrm{ m/s}$

$v_{tot_y} = 5.07 \textrm{ m/s}$

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This is College Physics Answers with Shaun Dychko. Our job in this question is to find the x and y components of this vector, the total vector, the resultant of <i>Va</i> plus <i>Vb</i>. It has components along the x axis, here's the <i>x</i> component of the vector and then here is the <i>y</i> component of the vector. Right here, this is <i>V total y</i> component, the amount by which the resultant goes up. Then along here horizontally, we have <i>V total x</i>, the amount by which this yellow vector goes to the right. To find the <i>x</i> component, we're going to need to know what this angle is in here and then we'll use cosine multiplied by the hypotenuse to find the length of this adjacent leg of this right triangle. So to find this angle we have to add the 26 and a half to the 22.5. That gives us 49 degrees and so the <i>x</i> component of the total then is the length of the vector, 6.72 meters per second, multiplied by cosine of 49. That gives 4.41 meters per second. The<i>y</i> component of the <i>V total</i> is going to be 6.72 meters per second multiplied by sine of 49 degrees, and we're using sine because this is the opposite leg of the triangle. We use sine always to get the opposite leg of a triangle, multiplied by the hypotenuse and that gives 5.07 meters per second.

## Comments

Submitted by xte11 on Sun, 06/23/2019 - 23:33

Submitted by ShaunDychko on Wed, 07/03/2019 - 20:21

In reply to I don't think this answer is… by xte11

Submitted by mkyee28@yahoo.com on Wed, 07/10/2019 - 11:48

Submitted by ShaunDychko on Mon, 07/15/2019 - 08:09

xandyaxes, we call these legsxandycomponentsalong these axes. All this is to say that the $49^\circ$ angle is the angle in the imaginary right triangle with legs parallel to thexandyaxes. It's the angle between the vector and the x-axis. Using $23^\circ$ would find components along different axes.In reply to how do you know to use 49… by mkyee28@yahoo.com

Submitted by barkzai2000 on Mon, 09/02/2019 - 10:47

Submitted by ShaunDychko on Tue, 09/03/2019 - 09:54

Nevertheless, since so many angles are given in the problem, you can figure out vector A and B magnitudes, you just can't do it with the resultant components. I would figure out the third angle inside the vector triangle (the three are supplementary), then use the sine law to find the lengths of vectors A and B.

In reply to How can you use this to find… by barkzai2000