Question

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?

Answer 1

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

Thus velocity in the y-direction is

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

and aceleration

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

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