2 kg નો બ્લોક 3.0 m/s ના સમક્ષિતિજ વેગથી

Displacement of body in $4 \,sec$ along $OE$

${s_x} = {v_x}t = 3 \times 4 = 12\;m$

Force along $OF$ (perpendicular to $OE$) $= 4\, N$

$\therefore {a_y} = \frac{F}{m} = \frac{4}{2} = 2\;m/{s^2}$

Displacement of body in $4\, sec$ along $OF$

$⇒$ ${s_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}$$ = \frac{1}{2} \times 2 \times {(4)^2} = 16\;m$ [As ${u_y} = 0$]

$\therefore $ Net displacement $s = \sqrt {s_x^2 + s_y^2} \; = \sqrt {{{(12)}^2} + {{(16)}^2}} = 20\;m$