Question

Find the components of along the x- and y-axes in Figure 3.57.

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Final Answer

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

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

Solution Video

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

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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 Va plus Vb. It has components along the x axis, here's the x component of the vector and then here is the y component of the vector. Right here, this is V total y component, the amount by which the resultant goes up. Then along here horizontally, we have V total x, the amount by which this yellow vector goes to the right. To find the x 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 x 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. They component of the V total 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

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.

In reply to by mkyee28@yahoo.com