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(C) Given: Initial velocity $u_x = 0$,$u_y = 0$. Acceleration $a_x = 6\,m/s^2$,$a_y = 8\,m/s^2$,and time $t = 4\,s$.Using the equation of motion $S =

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$80$

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(C) Given: Initial velocity $u_x = 0$,$u_y = 0$. Acceleration $a_x = 6\,m/s^2$,$a_y = 8\,m/s^2$,and time $t = 4\,s$.Using the equation of motion $S = ut + \frac{1}{2}at^2$:For the $x$-direction: $S_x = 0 + \frac{1}{2} 	imes 6 	imes (4)^2 = 3 	imes 16 = 48\,m$.For the $y$-direction: $S_y = 0 + \frac{1}{2} 	imes 8 	imes (4)^2 = 4 	imes 16 = 64\,m$.The distance from the origin is given by $S = \sqrt{S_x^2 + S_y^2}$.$S = \sqrt{48^2 + 64^2} = \sqrt{2304 + 4096} = \sqrt{6400} = 80\,m$.

$A$ body starts from rest from the origin with an acceleration of $6\,m/s^2$ along the $x$-axis and $8\,m/s^2$ along the $y$-axis. Its distance from the origin after $4\,s$ will be........$m$

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