Question

Sudhir walk .40 km in a direction 60 degrees west of north then goes .50km due west. what is his displacement?

Answer 1

here the two displacements are given as

d1 = 0.40 km in direction 60 degree west of north

so here we will have

$d_1 = 0.40sin60(- \hat i ) + 0.40 cos60 \hat j$

$d_1 = -0.35 \hat i + 0.20 \hat j$

other displacement is given as

d2 = 0.50 km due west

$d_2 = -0.50 \hat i$

total displacement is given as

$d = d_1 + d_2$

$d = -0.85 \hat i + 0.20 \hat j$

so the magnitude of the displacement is given as

$d = \sqrt{0.85^2 + 0.20^2}$

$d = 0.87 km$

and its direction is given as

$tan\theta = \frac{d_y}{d_x}$

$tan\theta = \frac{0.20}{0.85}$

$\theta = 13.2^0$

so displacement is 0.87 km towards 13.2 North of west

Answer 2

consider east-west direction along x-axis and north-south direction along y-axis

from the diagram

A = Ax i + By j  = - (0.40 Sin60) i + (0.40 Cos60) j = - 0.35 i + 0.20 j

B = Bx i + By j = - 0.50 i + 0 j

Net displacement is given as

D = A + B

D = (- 0.35 i + 0.20 j ) + (- 0.50 i + 0 j )

D = - 0.85 i + 0.2 j

magnitude of displacement is given as

|D| = sqrt((- 0.85)² + (0.2)²)

|D| = 0.87 km

direction of displacement is given as

θ = tan⁻¹(0.2/0.85)

θ = 13.5 deg north of west