You are here


You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -65.0 units.
(a) What are the two possibilities for its x component?
(b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the -x direction.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 
  1. $62.2\textrm{ units}$
  2. $156\textrm{ units, } 24.6^\circ \textrm{ above the negative x-axis}$

Please note that near the end of the video, the final answer to part b) says incorrectly that the vector $v_2$ is above the positive x-axis. $v_2$ is, in fact, above the negative x-axis, as shown in the drawing for part b).

Giancoli 7th Edition, Chapter 3, Problem 16


Chapter 3, Problem 16 is solved.

View sample solution


Handwriting could be better

Hi ianussery82, sorry if I was rushing through that one, but I hope it's legible enough to be helpful.

Thanks for the feedback.
Mr. Dychko

I am a little confused, in part A are we not solving for V^2? in which case we should solve for by square root of 90^2+(-65^2)? this comes out with the same answer posted but the math works out correctly.

Hi tthorne15, thanks for the question. If I don't quite understand your question, please just leave another comment. I think what you're asking about is the minus sign under the square root? Are you proposing $\sqrt{90^2 + 65^2}$ rather than the $\sqrt{90^2 - 65^2}$ shown in the video? The plus is used in the Pythagoras theorem when solving for the longest side of a right triangle (that side being known as the hypotenuse). In this case we're solving for one of the shorter sides (known as the "legs") instead, and after an algebraic manipulation (shown in the video) we end up with a minus.

All the best,
Mr. Dychko

I think the answer to this question is wrong. The coordinate of the resultant vector is (-142.2 , 65) which means that it is located in quadrant 2. Since the value of tangent is negative in quadrant 2 it makes sense as well. The angle is 24.56 degrees above the negative x - axis, not positive x-axis.

Hi chaegyunkang, thank you for spotting that, and for the thoughtful analysis. You're quite right! I'll change the written answer, and made a note about the video.
All the best with your studies,
Mr. Dychko