The x component of vector A is -25 m and the y component is 40m
a) what is the magnitude of vector A
b)What is the angle between the direction of vector A and the positive direction of x?
I got a) as 47.16m
by for b) I am having trouble I should be getting 122 degrees for the angle but I got -57.9 degrees which is wrong according to the book.
The angle will depend on how it is measured. Measuring from the y axis, the angle is
arctan(-25/40)
Now notice this from the above: I have indicated theta is the angle between y and the vector, NOT x and the vector.
To get the angle from the xaxis, add 90.
To find the magnitude of vector A, we can use the Pythagorean theorem. The magnitude of vector A, denoted as |A|, is given by:
|A| = sqrt(x^2 + y^2)
where x is the x-component of vector A (-25 m) and y is the y-component of vector A (40 m).
Substituting the values, we have:
|A| = sqrt((-25)^2 + (40)^2)
= sqrt(625 + 1600)
= sqrt(2225)
≈ 47.16 m
Therefore, the magnitude of vector A is approximately 47.16 m (rounding to two decimal places).
To find the angle between the direction of vector A and the positive direction of the x-axis, we need to use trigonometry. The angle, denoted as θ, can be found using the inverse tangent (arctan) function:
θ = arctan(y / x)
where x is the x-component of vector A (-25 m) and y is the y-component of vector A (40 m).
Substituting the values, we have:
θ = arctan(40 / -25)
Now, remember that the arctan function will give us the angle between the y-component and the x-component of a vector. To get the angle from the x-axis, we need to add 90 degrees.
θ = arctan(40 / -25) + 90
≈ -57.9 degrees
In your explanation, it seems that you forgot to add 90 degrees to the result obtained from the arctan function. Therefore, the correct angle between the direction of vector A and the positive direction of the x-axis is approximately -57.9 degrees. The negative sign indicates that the angle is measured in the clockwise direction from the positive x-axis.