A jogger travels a route that has two parts. The first is a displacement of 2.90...

A jogger travels a route that has two parts. The first is a displacement of 2.90 km due south, and the second involves a displacement that points due east. The resultant displacement + has a magnitude of 3.90 km. (a) What is the magnitude of , and (b) what is the direction of + as a positive angle relative to due south? Suppose that - had a magnitude of 3.90 km. (c) What then would be the magnitude of , and (d) what is the direction of - relative to due south?

Expert Solution

|
| 2.9 km
|
|______ b

I'm assuming the jogger is jogging km, not nautical miles. I suppose Jesus could jog nautical miles. Anywho

This is a right triangle and you know the hypotenuse and one leg, and want the other leg and then the angle between the vertical and the hypotenuse.

hyp^2 = 2.9^2 + b^2
3.9^2 = 2.9^2 + b^2
15.21 = 8.41 + b^2
b^2 = 15.21 - 8.41 =6.8
b = 2.60 km

The angle can be found through any of the reverse trig functions.

x = arcsin(2.6/3.9)
x = arcos(2.6/3.9)
x = arctan(2.6/2.9)

x = 41.8 degrees east of south

Subtracting a vector is the same as adding a negative vector. In this case, it would mean running west instead of east. The magnitude won't change. It will still be 2.6 km. The angle between the vector 3.9 km and the vector 2.9 km is now 41.8 degrees WEST of south. However, they want a vector relating to due west, so we have to find the complement.

90 - 41.8 = 48.1

So 48.1 degrees south of west.

genius_generous answered 6 months ago

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