Question

A man in search of his dog drives first 10 mi northeast, then 39 mi straight south, and finally 1 mi in a direction 30° north of west. What are the magnitude and direction of his resultant displacement?

Answer 1

In horizontal direction, we have

D_x = (10 mi) cos 45⁰ + (39 mi) cos 90⁰ - (1 mi) cos 30⁰

D_x = [(10 mi) (0.7071) + (39 mi) (0) - (1 mi) (0.8660)]

D_x = 6.205 mi

In vertical direction, we have

D_y = (10 mi) sin 45⁰ - (39 mi) sin 90⁰ + (1 mi) sin 30⁰

D_y = [(10 mi) (0.7071) - (39 mi) (1) + (1 mi) (0.5)]

D_y = - 31.429 mi

The magnitude of his resultant displacement which will be given by -

| D | = D_x² + D_y²

| D | = [(6.205 mi)² + (-31.429 mi)²]

| D | = 1026.284066 mi²

| D | = 32.03 mi

The direction of his resultant displacement which will be given by -

= tan^-1 (D_y / D_x)

= tan^-1 [(-31.429 mi) / (6.205 mi)]

= tan^-1 (-5.0651)

= - 78.8 degree

OR

= [360⁰ + (- 78.8)⁰]

= 281.2 degree

{ South of East }