Question

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. Part A What is the x component v⃗ x of v⃗ ?

Answer 1

`v_x = -6\ m/s`

Step-by-step explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

`v_x = v cos \theta`

velocity is in negative x-direction, v = -12 m/s

now,

`v_x = -12* cos 60^0`

`v_x = -12* 0.5`

`v_x = -6\ m/s`

Hence, the velocity x-component is equal to -6 m/s.