Consider a particle with initial velocity that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.
Is there a question in this?
The questions is x component of the vector?
To solve this problem, we can break down the initial velocity vector into its x and y components. The x component of the velocity will be directed along the negative x axis and the y component will be directed upwards.
Given:
Magnitude of the initial velocity (v) = 12.0 m/s
Angle (θ) = 60.0 degrees (measured above the negative x-axis)
To find the x and y components of the velocity, we can use trigonometry.
1. Finding the x-component:
Using the angle θ, we can find the x-component using cosine:
vx = v * cos(θ)
Substituting the given values:
vx = 12.0 m/s * cos(60.0 degrees)
Evaluating the expression:
vx = 12.0 m/s * 0.5
vx = 6.0 m/s
So, the x-component of the velocity is 6.0 m/s.
2. Finding the y-component:
Using the angle θ, we can find the y-component using sine:
vy = v * sin(θ)
Substituting the given values:
vy = 12.0 m/s * sin(60.0 degrees)
Evaluating the expression:
vy = 12.0 m/s * (√3 / 2)
vy = 6.0√3 m/s
So, the y-component of the velocity is 6.0√3 m/s.
Therefore, the particle's initial velocity can be expressed as a vector: vi = 6.0 m/s (-î) + 6.0√3 m/s (+ĵ), where î represents the unit vector along the x-axis and (+ĵ) represents the unit vector along the positive y-axis.