The x vector component of a displacement vector has a magnitude of 104 m and points along the negative x axis. The y vector component has a magnitude of 185 m and points along the negative y axis.
(a) Find the magnitude of .
________m
(b) Find the direction of .
_________° counterclockwise from the negative x axis
X = hor. = -104m.
Y = ver. = -185m.
a. D = sqrt(x^2 + y^2),
D = sqrt((-104)^2 + (-185)^2) = 212.2m.
b. tanAr = Y / X = -185 / -104=1.7788,
Ar = 60.7 deg.
A = 60.7 + 180 = 240.7 deg(Q3).
Ar = reference or related angle.
To find the magnitude of a displacement vector with given x and y components, you can use the Pythagorean theorem. The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of its two other sides.
In this case, the x and y components of the displacement vector form a right triangle. The magnitude of the displacement vector is the hypotenuse of this triangle.
(a) To find the magnitude of the displacement vector:
1. Square the x component: (-104 m)² = 10816 m²
2. Square the y component: (-185 m)² = 34225 m²
3. Add the squared x and y components: 10816 m² + 34225 m² = 45041 m²
4. Take the square root of the sum: √(45041 m²) = 212.46 m (rounded to two decimal places)
Therefore, the magnitude of the displacement vector is 212.46 m.
(b) To find the direction of the displacement vector:
1. Find the angle between the negative x-axis and the displacement vector.
2. Given that the x and y components of the displacement vector are negative, the vector points in the third quadrant of the coordinate plane.
3. To find the angle, use the inverse tangent function (arctan) with the absolute values of the components.
- arctan(abs(y component / x component))
- arctan(abs(-185 m / -104 m))
- arctan(185 m / 104 m)
- arctan(1.778)
4. Use a calculator to find the inverse tangent of 1.778. The result is approximately 59.88°.
5. Since the vector is in the third quadrant, the angle is measured counterclockwise from the negative x-axis.
6. Therefore, the direction of the displacement vector is approximately 59.88° counterclockwise from the negative x-axis.