The condition of the problem regarding the x-direction means xt′=4. We can immediately conclude xtt′′=0 and x=x0+4⋅t, where xo=x(t=0). This is uniform movement on x. Substitute this expression in y we find y=2cos(3x0+12t).
Thus velocity in the y-direction is
yt′=−2⋅12⋅sin(3x0+12t)=−24sin(3x0+12t)
and aceleration
ytt′′=−24⋅12⋅cos(3x0+12t)=−288cos(3x0+12t)
Answer: Acceleration along the axis x is 0. The acceleration a particle in y-direction −288cos(3x0+12t) where x0=x(t=0) particle position at the initial moment of time.
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