A jogger travels a route that has two parts. The first is a displacement of 2.20 km due south, and the second involves a displacement that points due east. The resultant displacement + has a magnitude of 4.10 km. (a) What is the magnitude of , and what is the direction of + as a positive angle relative to due south? (b) Suppose that - had a magnitude of 4.10 km. What then would be the magnitude of , and what is the direction of - relative to due south?
What Is Acceleration?
To solve this problem, we can use vector addition and trigonometry.
(a) Let's find the magnitude of the displacement vector + first. We can use the Pythagorean theorem to calculate it:
magnitude of + = √(2.20^2 + 4.10^2)
= √(4.84 + 16.81)
= √21.65
≈ 4.65 km
Now, to find the direction of +, we can use trigonometry. Since the first part of the route is due south, and the second part is due east, we have a right-angled triangle where the opposite side represents the displacement due south (2.20 km) and the adjacent side represents the displacement due east (4.10 km). Therefore, the angle relative to due south can be found using the arctan function:
angle = arctan(opposite/adjacent) = arctan(2.20/4.10)
Using a calculator, we can find that the angle is approximately 28.57 degrees.
So, the magnitude of + is approximately 4.65 km, and the direction of + as a positive angle relative to due south is approximately 28.57 degrees.
(b) If the displacement vector - has the same magnitude as + (4.10 km), the magnitude of - would still be 4.10 km.
The direction of - relative to due south would be the opposite of the direction of +. Since the direction of + is approximately 28.57 degrees relative to due south, the direction of - would be 180 degrees minus 28.57 degrees, which is approximately 151.43 degrees.
Therefore, the magnitude of - would be 4.10 km, and the direction of - relative to due south would be approximately 151.43 degrees.