Question

Suppose you first walk 5.5 m in a direction 20.0° west of north and then 18.5 m in a direction 40.0° south of west. (a) How far are you from your starting point in meters? (b) What is the angle, in degrees, of the compass direction of a line connecting your starting point to your final position measured south of west? (c) Repeat part (a), but for the second leg you walk 18.5 m in a direction 40.0° north of east. (d) Repeat part (a), but now you first walk 5.5 m in a direction 40.0° south of west and then 18.5 m in a direction 20.0° east of south.

Answer 1

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

Rx = Horizontal component = R*cos A

Ry = Vertical component = R*sin A

Using above rule:

Total displacement d = d1 + d2

by taking components,

|d1| = 5.5 m at 20 deg West of north = 5.5 m at 70 deg North of west (2nd quadrant, x < 0 & y > 0)

d1 = 5.5*cos 70 deg (-i) + 5.5*sin 70 deg (j)

d1 = (-1.88 i + 5.17 j) m

|d2| = 18.5 m at 40.0 deg South of West (3rd quadrant, x < 0 & y < 0)

d2 = 18.5*cos 40.0 deg (-i) + 18.5*sin 40.0 deg (-j)

d2 = (-14.17 i - 11.89 j) m

So, net displacement will be:

d = d1 + d2 = (-1.88 i + 5.17 j) + (-14.17 i - 11.89 j)

d = (-1.88 - 14.17) i + (5.17 - 11.89) j

d = -16.05 i - 6.72 j

then, distance from starting point(d) = sqrt (dx^2 + dy^2)

d = sqrt((-16.05)^2 + (-6.72)^2)

d = 17.4 m

b.)

Since x < 0 & y < 0, So d will be in 3rd quadrant

direction = = arctan(dy/dx) = arctan(-6.72/-16.05) = 22.7 deg below -ve x-axis

= 22.7 deg South of west.

Part C.

|d2| = 18.5 m at 40.0 deg North of East (1st quadrant, x > 0 & y > 0)

d2 = 18.5*cos 40.0 deg (i) + 18.5*sin 40.0 deg (j)

d2 = (14.17 i + 11.89 j) m

So, net displacement will be:

d = d1 + d2 = (-1.88 i + 5.17 j) + (14.17 i + 11.89 j)

d = (-1.88 + 14.17) i + (5.17 + 11.89) j

d = 12.29 i + 17.06 j

then, distance from starting point(d) = sqrt (dx^2 + dy^2)

d = sqrt((12.29)^2 + (17.06)^2)

d = 21.0 m

Part D.

|d1| = 5.5 m at 40.0 deg South of West (3rd quadrant, x < 0 & y < 0)

d1 = 5.5*cos 40.0 deg (-i) + 5.5*sin 40.0 deg (-j)

d1 = (-4.21 i - 3.535 j) m

|d2| = 18.5 m at 20 deg East of South = 18.5 m at 70 deg South of East (4th quadrant, x > 0 & y < 0)

d2 = 18.5*cos 70 deg (i) + 18.5*sin 70 deg (-j)

d2 = (6.33 i - 17.38 j) m

So, net displacement will be:

d = d1 + d2 = (-4.21 i - 3.535 j) + (6.33 i - 17.38 j)

d = (-4.21 + 6.33) i + (-3.535 - 17.38) j

d = 2.12 i - 20.915 j

then, distance from starting point(d) = sqrt (dx^2 + dy^2)

d = sqrt((2.12)^2 + (-20.915)^2)

d = 21.0 m