Question

The x component of vector A is -25.0m and the y component id +40.0m (a) what is the magnitude of A?(b) What is the angle between the direction of A and the positive direction of x?

Answer 1

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

$\vec A =$

Or also can be given by its polar components:

$\vec A =$

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

$r=\sqrt{x^2+y^2}$

$\displaystyle \theta=\arctan\frac{y}{x}$

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

$r=\sqrt{(-25)^2+40^2}$

$r=\sqrt{625+1600}$

$r=\sqrt{2225}$

$r\approx 47.2\ m$

(b)

$\displaystyle \theta=\arctan\frac{40}{-25}$

$\displaystyle \theta=\arctan (-1.6)$

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°