Question

In an effort to protect a rhino, volunteers are following its steps with air monitoring and ground cameras. The rhino starts on day 1 from ground camera A and walks for 1.5 km west along a straight line as seen from the air. The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B. In this location the rhino may find water and food, therefore the animal stays for the night. On the second day, the rhino moves 2.5 km directly south as recorded from the air by the volunteers. In these two days of tracking how far and in which direction does the rhino travel from camera A?

Answer 1

Answer:

Explanation:

We shall represent each displacement by vectors . i will represent east , -i west , j north and - j south .

Rhino walks 1.5 km west on day 1.

D₁ = - 1.5 i

The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B

D₂ = .7 sin15 i + .7cos15 j

On the second day, the rhino moves 2.5 km directly south

D₃ = - 2.5 j

D = D₁ + D₂ + D₃

= - 1.5 i + .7 sin15 i + .7cos15 j - 2.5 j

= - 1.5 i + .181 i + .676 j - 2.5 j

= - 1.32 i - 1.824 j

magnitude  of total displacement

= √ (1.32² +1.824²

= 2.25 km

For direction we shall calculate slope with x axis

Tanθ = - 1.824 / - 1.32

= 54°

So rhino will be towards 54° south of west as both x and y coordinates are negative.