To determine the magnitude of [tex]\( v_x \)[/tex] when the velocity [tex]\( v \)[/tex] makes an angle of [tex]\( 60^\circ \)[/tex] with the positive direction of the [tex]\( y \)[/tex]-axis, we will use trigonometric relationships.
Firstly, let's understand the given information:
- The velocity [tex]\( v \)[/tex] has a magnitude of [tex]\( 4.00 \)[/tex] meters/second.
- The angle [tex]\( \theta \)[/tex] between the velocity vector and the positive [tex]\( y \)[/tex]-axis is [tex]\( 60^\circ \)[/tex].
To find the magnitude of the [tex]\( x \)[/tex]-component of the velocity [tex]\( v_x \)[/tex], we can use the sine function. In this context, the sine function relates the angle [tex]\( \theta \)[/tex] to the ratio of the opposite side (which in this case will be [tex]\( v_x \)[/tex]) over the hypotenuse (which is [tex]\( v \)[/tex]).
The formula is:
[tex]\[
v_x = v \sin(\theta)
\][/tex]
Given:
[tex]\[
v = 4.00 \, \text{meters/second}
\][/tex]
[tex]\[
\theta = 60^\circ
\][/tex]
We need to convert the angle from degrees to radians:
[tex]\[
\theta \text{ (in radians)} = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \text{ radians} \approx 1.0472 \text{ radians}
\][/tex]
Next, we use the sine of [tex]\( 60^\circ \)[/tex]:
[tex]\[
\sin\left(\frac{\pi}{3}\right) = \sqrt{3} / 2 \approx 0.866 \quad \text{(approximately taking its value)}
\][/tex]
Now plug these values into the formula:
[tex]\[
v_x = 4.00 \times \sin\left(60^\circ\right) = 4.00 \times 0.866 \approx 3.46 \, \text{meters/second}
\][/tex]
Therefore, the magnitude of [tex]\( v_x \)[/tex] is approximately [tex]\( 3.46 \)[/tex] meters/second, which corresponds to option B.
So, the correct answer is:
[tex]\[ \boxed{3.46 \text{ meters/second}} \][/tex]
