Question

Suppose you first walk 10.5 m in a direction 20° west of north and then 25 m in a direction 40° south of west as shown in the figure.

a. What is the component of your displacement in the x-direction?

b. What is the component of your displacement in the y-direction?

c. How far are you from your starting point?

d. What is the angle of a line connecting your starting position to your final position, measured South of West, in degrees?

Answer 1

Suppose given that Vector is R and it makes angle with +x-axis, then it's components are given by:

Rx = R*cos

Ry = R*sin

Using above rule:

A = 10.5 m, direction 20 deg W of N = 10.5 m, direction 110 deg N of E (in +x-axis)

Ax = 10.5*cos 110 deg = -3.59 m

Ay = 10.5*sin 110 deg = 9.87 m

B = 25 m, direction 40 deg S of W = 25 m, direction 220 deg N of E (in +x-axis)

Bx = 25*cos 220 deg = -19.15 m

By = 25*sin 220 deg = -16.07 m

Now

R = Rx + Ry = A + B

R = (Ax + Bx) i + (Ay + By) j

R = (-3.59 + (-19.15)) i + (9.87 + (-16.07)) j

R = -22.74 i - 6.20 j

Part-A.

So Component of displacement in x-direction will be

Rx = -22.74 m

Part B.

So Component of displacement in y-direction will be

Ry = -6.20 m

Part C.

So distance R will be

|R| = sqrt (Rx^2 + Ry^2)

|R| = sqrt ((-22.74)^2 + (-6.20)^2)

|R| = 23.57 m

Part D.

Direction will be given by:

theta = arctan (Ry/Rx)

theta = arctan (-6.20/-22.74)

theta = 15.25 deg below -ve x-axis = 15.25 deg South of west

Let me know if you've any query.