To solve this problem, let's consider the situation where the velocity vector [tex]\( v \)[/tex] is equally inclined to the negative directions of both the [tex]\( x \)[/tex]-axis and the [tex]\( y \)[/tex]-axis.
The velocity vector [tex]\( \vec{v} \)[/tex] can be resolved into components along the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] axes. Since it is equally inclined, the angle with each axis is [tex]\( 45^\circ \)[/tex].
Given:
[tex]\[ v = 5.00 \ \text{meters/second} \][/tex]
1. We know the vector is equally inclined to both axes, meaning it makes an angle of [tex]\( 45^\circ \)[/tex] with each axis. In such cases, the components of [tex]\( v \)[/tex] along the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] directions ([tex]\( v_x \)[/tex] and [tex]\( v_y \)[/tex]) will have equal magnitudes but negative signs because they are in the negative directions.
2. The components of the velocity vector [tex]\( v \)[/tex] along the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] axes can be found using trigonometric relations for a [tex]\( 45^\circ \)[/tex] angle:
[tex]\[ v_x = v \cos(45^\circ) \][/tex]
[tex]\[ v_y = v \cos(45^\circ) \][/tex]
Since [tex]\( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)[/tex]:
[tex]\[ v_x = v \cdot \frac{1}{\sqrt{2}} \][/tex]
[tex]\[ v_y = v \cdot \frac{1}{\sqrt{2}} \][/tex]
3. Substituting the value of [tex]\( v = 5.00 \ \text{meters/second} \)[/tex]:
[tex]\[ v_y = 5.00 \cdot \frac{1}{\sqrt{2}} \][/tex]
[tex]\[ v_y = 5.00 \cdot \frac{\sqrt{2}}{2} \][/tex]
Simplifying the expression:
[tex]\[ v_y = 5.00 \cdot \frac{\sqrt{2}}{2} = 5.00 \cdot 0.7071 = 3.5355339059327378 \][/tex]
Since the vector is in the negative [tex]\( y \)[/tex]-direction:
[tex]\[ v_y = -3.5355339059327378 \ \text{meters/second} \][/tex]
Hence, the correct answer is:
A. -3.53 meters/second
