The $$x$$ and $$y$$ coordinates of the particle at any time are $$x = 5t - 2{t^2}$$  and $$y = 10t$$  respectively, where $$x$$ and $$y$$ are in metres and $$t$$ in seconds. The acceleration of the particle at $$t = 2\,s$$  is

Solution :
$$\eqalign{ & {\text{Given,}}\,x = 5t - 2{t^2} \cr & {\text{Velocity of the particle,}} \cr & {v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {5t - 2{t^2}} \right) = 5 - 4t \cr & {\text{Acceleration, }}{a_x} = \frac{d}{{dt}}{v_x} = - 4\;m{s^{ - 2}} \cr & {\text{Also}}\,y = 10t \cr & {\text{Velocity,}}\,{v_y} = \frac{{dy}}{{dt}} = 10 \cr & \therefore {\text{Acceleration }}{a_y} = \frac{{d{v_y}}}{{dt}} = 0 \cr & \therefore {\text{Net acceleration of the particle,}} \cr & {a_{net}} = {a_x}\widehat i + {a_y}\widehat j = \left( { - 4\;m{s^{ - 2}}} \right)\widehat i \cr & {\text{or}}\,{a_{net}} = - 4\,\widehat i\,m{s^{ - 2}} \cr} $$