If the path of a moving point is the curve x | vedclass

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(C) Given the position coordinates of the moving point are $x = at$ and $y = b \sin(at)$.First,we find the velocity components by differentiating with

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Varies as the distance from the axis of $y$

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(C) Given the position coordinates of the moving point are $x = at$ and $y = b \sin(at)$.First,we find the velocity components by differentiating with respect to time $t$:$v_x = \frac{dx}{dt} = a$$v_y = \frac{dy}{dt} = ab \cos(at)$Next,we find the acceleration components by differentiating the velocity components with respect to time $t$:$a_x = \frac{d^2x}{dt^2} = 0$$a_y = \frac{d^2y}{dt^2} = -ab^2 \sin(at)$Since $y = b \sin(at)$,we can substitute this into the expression for $a_y$:$a_y = -a^2(b \sin(at)) = -a^2y$Thus,the acceleration vector is $\vec{a} = (0, -a^2y)$.This shows that the acceleration is directed along the $y$-axis and its magnitude varies directly as the distance from the $x$-axis (which is the value of $y$).

If the path of a moving point is the curve $x = at$,$y = b \sin(at)$,then its acceleration at any instant

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