Question
Find the components of $v_\textrm{tot}$ along the $x$ and $y$ axes in Figure 3.55.
Question Image

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

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

Solution Video

# OpenStax College Physics Solution, Chapter 3, Problem 11 (Problems & Exercises) (1:43) View sample solution

## Calculator Screenshots Video Transcript

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.

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

I don't think this answer is right

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

how do you know to use 49 degrees for both the x and y components? why not the 23 degrees?

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

When finding components we want to imagine a right triangle (a triangle with a $90^\circ$ angle, in other words) that has the vector as it's hypotenuse (which is the name of the longest side opposite the $90^\circ$ angle), and the components are the legs. When the legs of the right triangle are parallel to the x and y axes, we call these legs x and y components along these axes. All this is to say that the $49^\circ$ angle is the angle in the imaginary right triangle with legs parallel to the x and y axes. It's the angle between the vector and the x-axis. Using $23^\circ$ would find components along different axes.

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

How can you use this to find the magnitudes of velocities for vector A and vector B?

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

Knowing the components of the resultant isn't actually useful for finding the magnitudes of vectors A nor B. This is because vectors A and B could have a variety of different lengths, but with adjusted angles, they could nevertheless create the same resultant.
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.

Submitted by jessicamatt on Wed, 03/11/2020 - 15:04

I'm not coming up with the same answer when putting the equation in?

Submitted by ShaunDychko on Wed, 03/11/2020 - 21:07

Keep in mind that we're finding the components of the resultant vector (it's a bit confusing that the picture has three vectors on it, so let's be sure we're looking at the right one. It's labeled $v_{\textrm{tot}}$). Have you looked at the calculator screenshot showing how I plugged this into the calculator? Are you seeing anything different on your calculator, or something amiss with how I've plugged values into mine?