Question

1. A hiker travels 2 km due east of his starting point. Then he travels 1 km northwest. Finally, he travels 3 km due north.

1. How far is the hiker from his starting point after the 3 displacements and in what direction?
2. Draw the three displacement vectors to scale and add them graphically

1. Find the magnitude and direction of A, B and A + B for A = -4i – 7j and B = 3i – 2j .

1. Describe the following vectors using the unit vectors i and j.

1. A velocity of 10 m/s at an angle of elevation of 60 degrees.
2. A vector A of magnitude A = 5m and θ = 225 degrees.
3. A displacement from the origin to the point x = 14 m and y = -6 m.

1. A stationary radar operator determines that a ship is 10 km south of him. An hour later the same ship is 20 km southeast of him. If the ship moved at a constant velocity and always in the same direction. What was its average velocity during this time in both magnitude and direction?

Answer 1

In the below figure

Initially hiker travels 2km in east

Z 3 km 1km w 2 km SR

When hiker travels in North-west direction for 1 km making an angle 45 degree with the horizontal

So there will be two components

Horizontal components rcos$\theta$ = 1*cos45 = 1/ √2

Vertical components rsin$\theta$ = 1*sin45 = 1/ √2

Position of hiker after 2nd displacement due east   = 2- rcos$\theta$ = 2-(1/ √2) = 1.29km

Position of hiker after 2nd displacement due north = rsin$\theta$ = 1*1/ √2 = 1/ √2 = 0.707 km

after 3rd displacement

horizontal components dx = 1.29 km

vertical components dy =3+ 0.707= 3.707 km

Magnitude of total displacement r = Sqrt (dx^2 + dy^2)

r = Sqrt (1.29^2+3.707^2)

r = 3.93 km

this direction will be towards north-east from initial point.

Magnitude and direction of

A, B and A + B for A = -4i – 7j and B = 3i – 2j

A+B = (-4i – 7j )+ (3i-2j )

A+B = -i -9j

Direction tan$\theta$ = y/x

$\theta$ = tan^-1(y/x)

$\theta$ = tan^-1(-9/-1) = tan^-1(9)

$\theta$ = 83.66 Degree