To find the magnitude of the vector with components \[ x = -309 \, \text{m} \] and \[ y = 187 \, \text{m} \], we can use the Pythagorean theorem. The magnitude \( |\mathbf{v}| \) of the vector \(\mathbf{v}\) with components \( (x, y) \) is given by:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]
Substitute the given values:
\[ |\mathbf{v}| = \sqrt{(-309)^2 + 187^2} \]
Calculate the squares:
\[ (-309)^2 = 95581 \]
\[ 187^2 = 34969 \]
Sum the squares:
\[ 95581 + 34969 = 130550 \]
Take the square root of the sum:
\[ |\mathbf{v}| = \sqrt{130550} \approx 361.34 \, \text{m} \]
To find the direction (angle) \( \theta \) of the vector, we can use the inverse tangent function. The angle \(\theta\) (measured counterclockwise from the positive x-axis) is given by:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
Substitute the given values:
\[ \theta = \tan^{-1}\left(\frac{187}{-309}\right) \]
\[ \theta = \tan^{-1}\left(-0.6052\right) \]
Calculate the angle using the arctangent function:
\[ \theta \approx -31.54^\circ \]
Since the vector is in the second quadrant (negative x and positive y components), we must add 180° to the angle to get the correct direction:
\[ \theta = -31.54^\circ + 180^\circ \]
\[ \theta \approx 148.46^\circ \]
Therefore, the magnitude of the vector is approximately \( 361.34 \, \text{m} \) and its direction is approximately \( 148.46^\circ \).
A vector has an x-component of -309 m, and a y- component of 187m.
Find the magnitude of the vector [?] degrees
1 answer