Question

I start walking. The 1^st leg of my trip I walk d_A = 65 m at θ_A = 23° south of east. The 2^nd leg of my trip I walk d_B = 135 m at θ_B = 18° north of east. On my final leg I walk d_C = 95 m at θ_C = 53° north of west. Choose the coordinate system so that x is directed towards the east, and y is directed towards the north. Part (a) Write an expression for the x-component of the final displacement in terms of the given quantities. (b) Write an expression for the y-component of the final displacement in terms of the given quantities. (c) What is the magnitude of my displacement vector (in meters) as measured from the origin? (d) What is the angle of my displacement vector as measured counterclockwise from the +x-axis?

Answer 1

Please Note: All the final answer of each part are in bold words.

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

Now we need to find A + B + C

Using above rule:

A = 65 m, direction 23 deg S of E (4th quadrant, x > 0 & y < 0)

Ax = 65*cos 23 deg = 59.83 m

Ay = -65*sin 23 deg = -25.40 m

B = 135, direction 18 deg N of E (1st quadrant, x > 0 and y > 0)

Bx = 135*cos 18 deg = 128.39 m

By = 135*sin 18 deg = 41.72 m

C = 95 m, direction 53 deg N of W (2nd quadrant, x < 0, & y > 0)

Cx = -95*cos 53 deg = -57.17 m

Cy = 95*sin 53 deg = 75.87 m

Now

D = A + B + C

D = (Ax + Bx + Cx) i + (Ay + By + Cy) j

D = (59.83 + 128.39 - 57.17) i + (-25.40 + 41.72 + 75.87) j

D = 131.05 i + 92.19 j

Part (a)

As calculated from above expression of x-component of the final displacement will be given by:

D_x = d_A*cos _A + d_B*cos _B - d_C*cos _C

Using above calculated values

x-component of the final displacement = 131.05 m

Part (b)

As calculated from above expression of y-component of the final displacement will be given by:

D_y = -d_A*sin _A + d_B*sin _B + d_C*sin _C

Using above calculated values

y-component of the final displacement = 92.19 m

Part (c)

So Magnitude of D will be

|D| = sqrt (Dx^2 + Dy^2)

|D| = sqrt (131.05^2 + 92.19^2)

|D| = 160.2 m

Part (d)

Direction will be given by:

theta = arctan (Dy/Dx)

theta = arctan (92.19/131.05)

theta = 35.12 deg = 35.12 deg angle with CCW with the +ve x axis

Since x > 0 and y > 0, then vector will be in 1st quadrant