`v_x=(dx)/(dt)=-6\ sin\ 3t`

`v_y=(dy)/(dt)=-2\ sin\ 2t`

At `t = 4.1`, `v_x = 1.579` and `v_y = -1.88`.

So

`v=sqrt((v_x)^2+(v_y)^2)`

`=sqrt((1.579)^2+(-1.88)^2)`

`=2.46\ "cm"//s`

For velocity, we need to also indicate direction. First, we find the appropriate acute angle (the "reference" angle):

`alpha=tan^-1(1.88 / 1.579)=50^@`

So since we are in the 4th quadrant when `v_x` is positive and `v_y` is negative, the required angle is `360^@ - 50^@ = 310^@`.

(See the following for background on how to find this angle: Trigonometric Functions of any Angle.)

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