Answer to Question #219165 in Mechanics | Relativity for Maxmiliano Onai

Question

a. In the sum A+B = C, vector A has a magnitude of 12.0m and is angled 40.0 degrees counterclockwise from the +x direction, and vector C has magnitude of 15.0m and is angled 20.0 degrees counterclockwise from
the −x direction. What are
i. the magnitude?
ii. the angle (relative to +x) of B?

b. An electron moving along the x axis has a position given by x = (16te exponent -t )m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

c. A vector B, when added to the vector C = 3.0i+4.0j, yields a resultant vector that is in the positive y direction and has a magnitude equal to that of C. What is the magnitude of B?

Answer 1

$A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m$

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

$C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m$

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx = cx and ay + by = cy. From this two equations we can obtain B.

$b_x= (15*cos(200°) - 12*cos(40°)) m = -23.288 m$

$b_y = (15*sin(200°) - 12*sin(40°)) m = -12.843 m$

Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.

Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.$b_x = B'*cos(208.8°)$ so B' = (-23.288

m)/cos(208.8°) = 26.58 m.

So the magnitude of B is 26.58 m