Question #99803
a particle moves along a path y=2cos3x in such a way that the component of its velocity in the x-direction is 4 units. find its acceleration?
1
Expert's answer
2019-12-03T10:43:08-0500

The condition of the problem regarding the x-direction means xt=4x^{'}_t=4. We can immediately conclude xtt=0x^{''}_{tt}=0 and x=x0+4tx=x_0+4\cdot{t}, where xo=x(t=0)x_o=x(t=0). This is uniform movement on x. Substitute this expression in y we find y=2cos(3x0+12t)y=2cos(3x_0+12t).

Thus velocity in the y-direction is

yt=212sin(3x0+12t)=24sin(3x0+12t)y^{'}_t=-2\cdot{12}\cdot{sin(3x_0+12t)}=-24sin(3x_0+12t)

and aceleration

ytt=2412cos(3x0+12t)=288cos(3x0+12t)y^{''}_{tt}=-24\cdot{12}\cdot{cos(3x_0+12t)}=-288cos(3x_0+12t)

Answer: Acceleration along the axis x is 0. The acceleration a particle in y-direction 288cos(3x0+12t)-288cos(3x_0+12t) where x0=x(t=0)x_0=x(t=0) particle position at the initial moment of time.


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