A body of mass 2 kg has an initial velocity of 3 meters per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 seconds will be
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Displacement of body in 4 sec along OE
\(s_{x} = v_{x} t = 3 \times 4 = 12 \text{ m}\)
Force along OF (perpendicular to OE ) = 4N
\(\therefore a_y = \frac{F}{m} = \frac{4}{2} = 2 \, \mathrm{m/s^2} \\ \text{Displacement of body in 4 sec along OF} \\ \Rightarrow s_y = u_y t + \frac{1}{2} a_y t^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \mathrm{m} \, [\text{As } u_y = 0]\)
\(\cdot\) Net displacement \(s = \sqrt{s_x^2 + s_y^2} = \sqrt{(12)^2 + (16)^2} = 20 \text{ m}\)
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