I start walking. The 1^st leg of my trip I walk d_A = 150 m at theta_A = 6 degre

Question

I start walking. The 1^st leg of my trip I walk d_A = 150 m at theta_A = 6 degree south of east. The 2^nd leg of my trip I walk d_B = 105 m at theta_B = 28 degree north of east. On my final leg I walk d_C = 75 m at theta_C = 69.5 degree north of west. Choose the coordinate system so that x is directed towards the east, and y is directed towards the north. Write an expression for the x-component of the final displacement in terms of the given quantities. R_x = _______ Write an expression for the y-component of the final displacement in terms of the given quantities. What is the magnitude of my displacement vector (in meters) as measured from the origin? What is the angle of my displacement vector as measured counterclockwise from the +x-axis?

Explanation / Answer

a)Let this be Rx.

Rx = dA cos(theta-a) + dB cos(theta-b) + dc cos(theta-c)

Rx = 150 x cos6 + 105 x cos28 + 75 x cos69.5 = 268.15 m

Hence, Rx = 268.15 m

b)Ry will be:

Ry = dA sin(theta-a) + dB sin(theta-b) + dC sin(theta-c)

Ry = 150 x sin6 + 105 x sin28 + 75 x sin69.5 = 135.22 m

Ry = 135.22 m

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

R = sqrt (268.15^2 + 135.22^2) = 300.31 m

Hence, R = 300.31 m

d)theta = tan^-1 (135.22/268.15) = 26.76 deg

Hence, theta = 26.76 deg

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---