What is the acceleration of the point at this instant?

Equation of trajectory:-

$ y=\dfrac {x^2}3 -----------(i)$

Differentiating w.r.t time,

$ \dfrac {dy}{dt}=2/3 \times \dfrac { dx}{dt} ------------(ii)$

Differentiating w.r.t time,

$ \dfrac {d^2 y}{dt^2}=\dfrac 23 (\dfrac {dx}{dt})^2+ \dfrac 23 \times \dfrac {d^2 x}{dt^2} ----------(iii) $

At x=3, equation (ii) becomes:-

$ \dfrac {dy}{dx}=\dfrac {2×3}3 \dfrac {dx}{dt}\\ \dfrac {dy}{dx}=\dfrac {2dx}{dt} ------------(iv)$

Given that the resultant speed is constant,

$v=\sqrt{(\dfrac {dy}{dx})^2+(\dfrac {dx}{dt})^2}=8 ---------(v) \\ \therefore 4 (\dfrac {dy}{dx})^2+(\dfrac {dx}{dt})^2=64 \\ V_x= \dfrac {dx}{dt}=3.578m/s \Rightarrow Ans \\ \text {And } V_y= \dfrac {dy}{dt}=7.155m/s Ans[from (v)] \\ \text {From (v) }\\ (\dfrac {dy}{dx})^2+(\dfrac {dx}{dt})^2=8^2 i.e \space \space V_y^2+V_x^2=8^2 $

Differentiating w.r.t time,

$ 2V_y \dfrac {dV_y}{dt} + 2V_x \dfrac {dV_x}{dt}=0 \\ \therefore dy=-\dfrac {V_x}{V_y} a_x------------(vi)$

Substituting (vi) in (iii) we get,

$ -\dfrac {V_x}{V_y} a_x=\dfrac23 V_x^2 + \dfrac 23 xa_x $

At $x=3, V_x=3.578m/s \space\space\space V_y=7.155m/s \\ \therefore -\dfrac 12 a_x = \dfrac 23(3.578)^2 + \dfrac 23 × 3a_x \\ \therefore a_x=-3.414m/s^2 \Rightarrow Ans \\ \therefore a_y=1.707m/s^2 \Rightarrow Ans \\ \text {Resultant } a=3.817 m/s^2 \space \space 26.565⁰ $

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