Physics Displacement Problem?

Question

The pronghorn antelope, found in North America, is the best long-distance runner among mammals. It has been observed to travel at an average speed of more than 55 km/h over a distance of 6.0 km. Suppose the antelope runs a distance of 5.0 km in a direction 11.5 degrees north of east, turns, and then runs 1.0 km south. Calculate the resultant displacement.

Answer 1

The initial poster didn't read the entire question. This is not as simple as the pythagorean theorem idea he/she had. You need to use vectors to solve this problem. Assume X is east-west and Y is north-south, with positive being east and north.

The first vector is 5.0 km at 11.5 degrees north of due east. You need to find the X and Y components of this vector. Use SOH CAH TOA, and you have the H, and you have the angle. So to get the value of the X component, you need to use sin:

cos (11.5) = X/5.0, so the X component is 5 cos (11.5)

similarly for the Y component, use sin:

sin (11.5) = Y/5.0, the Y component is 5 sin (11.5)

Great, so the vector for the first distance is
<5 cos(11.5), 5 sin(11.5)>

The second distance is 1.0 km due south. Well, this is only in the Y direction, so the vector would be <0, -1> Use the negative sign because we assigned north as positive, and south as negative. Now we can add the components of the vectors to get a resultant vector.

<5 cos(11.5), 5 sin(11.5) -1>

This says that the resultant vector has an X component of
5 cos(11.5) and a Y component of 5 sin(11.5) -1.
Using the Pythagorean theorem, the magnitude of this resultant vector would be
sqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ]

Now to get the direction, we need to use trigonometry again. We have the opposite, and the adjacent, which is TOA.

tan(theta) = Y/X = (5 sin(11.5)-1)/(5 cos(11.5))

Take arctan of both sides:
theta = arctan[(5 sin(11.5)-1)/(5 cos(11.5))]

So the answer?
The resultant displacement vector is:

sqrt [ (5 cos (11.5))^2 + (5 sin(11.5)-1)^2 ] at an angle of
arctan[(5 sin(11.5)-1)/(5 cos(11.5))]

Answer 2

Assume (0,0) be the point from which antelope starts. Draw a diagram and see that the final coordinates of the antelope are:


(5 cosine(11.5), [5 sine(11.5) -1]), so the antelope is

at a distance 's' from the origin at an angle x degrees north of east given by: let y 11.5 degree

s = sqrt(25 [cos y][cos y] + 25 [sin y][csin y] +1 - 10sin y)
= sqrt(26 - 10 siny), and
tan x = [5 sine(11.5) -1]/5 cosine(11.5)
= tan 11.5 - (1/5) sec 11.5

I do not want to edit my answer. I want to tell Brian that he should draw the diagram and see for himself that he should have added the two vectors by triangle law and not by parallelogram law.