what is the compass direction of the line connecting your starting point to your final position

Question

Suppose you first walk 12.0 k in a management due west of north and and so 20.0 m in a direction $40.0^\circ$ due south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you correspond the ii legs of the walk as vector displacements A and B , every bit in Effigy 3.54, then this problem finds their sum R = A + B.)

$nineteen.5 \textrm{ 1000, } iv.65^\circ \textrm{ S of W}$

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 3, Problem 5 (Problems & Exercises) (v:01)

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Video Transcript

This is College Physics Answers with Shaun Dychko. And then it'southward nice that this textbook provides u.s.a. with a nice picture of our situation here. First yous walk 12 meters in a direction 20 degrees to the due west of north, and and then walk 20 meters in the direction forty degrees southward of due west and you end up here. The resultant vector or the displacement vector connecting your initial position with your final position is here. These vectors A and B have been added together using the head to tail method. Our chore is to figure out what is the size of this resultant vector and what is its direction. So we know that the x component of the resulting vector volition exist the sum of the x components of the other vectors, Ax and Bx. So to calculate them, detect first of all that there is a bit of a tricky matter here that we apply sine on the ane manus and so nosotros use cosine on the other. So you can't merely follow a recipe where you lot memorize that the 10 component is always cosine and the y component is always sine. That's not necessarily true. Information technology often is, simply in this item question, though the way they've drawn it and the angles that they've given you, it turns out that the x component of the A vector is using sine because it's the opposite leg of this triangle hither. And so this is the triangle that makes upwardly vector A and its opposite leg is opposite the angle given and then to observe the opposite leg of the triangle nosotros use the trigonometric function sine, multiplied past the hypotenuse. So we're finding the side by side here which volition be the y component of A, we're going to end up using cosine. Okay. So we go 12 meters times sine twenty and nosotros put a negative in front of it because we tin see that it is directed to the left. Nosotros'll assume that to the right is positive, and then we have a minus in front of that. And so too, minus 20 meters magnitude of vector B, multiplied by cosine of forty because the x component of vector B is the next leg of this triangle, and this gives us negative 19.43 meters as the x component of the resultant. At present, for the y component nosotros have 12 meter magnitude of vector A, times by the cosine of 20 to get the adjacent leg of this triangle, that'southward positive and so we did non put a minus in front of -- we did non put a minus in forepart of this term. But nosotros do put a minus in front end of the y component of vector B because it is directed downwards. Here is the y component of vector B directed down here. So nosotros accept a minus 20 meters times sine twoscore, and that gives negative 1.579 meters and that'due south the resultant y component. Then if we were to draw the resultant y component it would be like this. What I hateful to say is I draw the components of the resulting vector it would look like this, this beingness the x component of the resultant, negative nineteen.43, and then the y component directed down, negative 1.579, and this is the resultant here. Okay. And so now that we know the legs of this triangle, nosotros can now calculate the length of the hypotenuse which is the magnitude of the resultant, and so we can also find the angle in hither. And then the size of the resultant is the square root of the sum of the squares of the components, and so that's the square root of negative 19.43 meters squared, plus negative 1.579 meters squared, giving the states xix.5 meters is the length of the resultant. Then for the angle, we take the inverse tangent of its y component divided past its x component and in that location is little vertical confined here to say absolute value or to say ignore the negative sign. We're just going to take the magnitude of these components. Then the direction we'll effigy out by looking at the pic. Nosotros don't need positives or negatives to tell us direction. Then, we take 4.65 degrees considering that's a picayune flake in here, this is four.65 degrees towards the due south compared to west. We're south of w. So our resultant is 19.v meters, iv.65 degrees south of west.

Solutions for problems in chapter iii

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