To calculate the vector magnitude of acceleration:
\[ A = \sqrt{A_x^2 + A_y^2} \]
To calculate the angle of the acceleration vector:
\[ \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \]
Where:
Vector acceleration is a measure of how quickly the velocity of an object is changing in both the x and y directions. It is expressed as a vector quantity, meaning it has both magnitude and direction. The magnitude of vector acceleration represents the overall rate of change in velocity, while the angle describes the direction of this change in relation to the x-axis. In the International System of Units (SI), the unit for vector acceleration is meters per second squared (m/s²). The angle can be expressed in either degrees or radians.
Let's assume the following values:
Using the formulas:
\[ A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s²} \]
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \]
The vector magnitude of acceleration is 5 m/s², and the angle of the acceleration vector is approximately 53.13 degrees.
Let's assume the following values:
Using the formulas:
\[ A = \sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41 \text{ m/s²} \]
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = 45^\circ \]
The vector magnitude of acceleration is approximately 1.41 m/s², and the angle of the acceleration vector is 45 degrees.