dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of the vector that will take the dolphin back to its starting point? (Enter the magnitude in km and the direction in degrees north of east.) Hint

Explanation:

We shall represent all the displacement in terms of vector . We shall consider east as i , and north as j . south of west direction will be represented by - i - j .

displacement of 1.9 km due east

D₁ = 1.9 i

vector representing south of west = - i - j

unit vector = - i - j / √ 2

7.2 km south of west = 7.2 ( - i - j ) / √ 2

D₂ = - 5.09 ( i + j )

Total displacement

= D₁ + D₂

= 1.9 i  - 5.09 ( I + J )

D = - 3.19 i - 5.09 j .

magnitude of D = √ ( 3.19² + 5.09² )

= 6 km .

Direction of D

Tanθ = 5.09 / 3.19 = 1.59

θ = 58°

So direction will be 58° south of west .

To reach the starting point , he shall have to go in opposite direction .

So he shall have to go in the direction of north of east at angle 58° by a displacement of 6 km .

this would be about 43%