Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis. 1.What is the x component vx of v⃗ ? 2.What is the y component vy of v⃗ ?

Answer :

skyluke89

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

[tex]v_x = v cos \theta[/tex]

[tex]v_y = v sin \theta[/tex]

where

v = 12.0 m/s is the magnitude of the vector

[tex]\theta[/tex] is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

[tex]\theta=180^{\circ} - 60^{\circ} = 120^{\circ}[/tex]

So, the two components are:

[tex]v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s[/tex]

[tex]v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s[/tex]

The change of displacement with respect to time is defined as the velocity. The x component of the velocity will be -6.0 m/sec. while the velocity in the y-direction will be 10.4 m/sec.

What is velocity?

The change of displacement with respect to time is defined as the velocity.

velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

[tex]V_x = V {cos\theta}[/tex]

[tex]V_y = V {sin\theta}[/tex]

[tex]\theta[/tex] = angle measured from the positive x-axis

In the above conditions, the angle is given from the - ve x-axis is 60 degrees. If it is measured from the + ve x-axis it will be 180- 60= 120 degrees.

Given velocity v = 12.0 m/sec

[tex]V_x = 12{cos20}[/tex] = -6.0 m/sec

[tex]V_y = 12 {sin120}[/tex] = 10.4 m/sec

Hence the x component of the velocity will be -6.0 m/sec. while the velocity in the y-direction will be 10.4 m/sec.

To learn more about the velocity refer to the link ;

https://brainly.com/question/862972

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