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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. What is the x component v⃗ x of v⃗ ? What is the y component v⃗ y of v⃗ ?
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I know I use the COS (60) to find the x value but the "negative x axis part is throwing me off...
60 degrees above the negative x axis is equal to 120 degrees.
|dw:1383008419858:dw|
With forces you have a magnitude (strength) and a direction. So you can use the 60 degrees and add the negative sign in afterwards. \[v_{x} = 12cos(60) [m/s] -x direction\] or you can write as \[v_{x} = -12cos(60) [m/s] \] Simular for the y-direction
The negative sign just tells you direction about the velocity
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