A teacher sends her students on a treasure hunt. She gives the following instructions:

1. Walk 300 m north

2. Walk 400 m northwest

3. Walk 700 m east-southeast and the treasure is buried there.

As all the other students walk off following the instructions, Joe physics student quickly adds the displacements and walks in a straight line to find the treasure. How far and in what direction does Joe need to walk?

A) 399 m in a direction 52.5° north of east

B) 187 m in a direction 67.3° north of east

C) 284 m in a direction 28.2° west of north

D) 481 m in a direction 40.9° north of east

E) The treasure position cannot be reached in one straight walk.

D) Joe need to walk 481 m in a direction 40.9° north of east

Explanation:

Given:

1. Walk 300 m north

2. Walk 400 m northwest

3. Walk 700 m east-southeast and the treasure is buried there

Taking north as positive y axis and east as positive x axis.

Resolving their positions into x and y vector components.

1. d1 = 300j

2. d2 = - 400cos45i + 400sin45j

3 d3 = 700cos22.5i - 700sin22.5j

Resultant position.

d = d1+d2+d3

d = (-400cos45 + 700cos22.5) i + (300+400sin45-700sin22.5) j

d = (363.873) i + (314.964) j

D = √ ((363.873) ^2 + (314.964) ^2)

D = 481.25m ~ = 481m

Angle = taninverse (314.964/363.873)

Angle = 40.9°

Since the x and y components are both positive, it implies that their position is north of east

Therefore, Joe need to walk 481 m in a direction 40.9° north of east