A body of mass $2 ext{ kg}$ has an initial velocity of $3 ext{ m/s}$ along $OE$ and it is subjected to a f

Question

A body of mass $2 	ext{ kg}$ has an initial velocity of $3 	ext{ m/s}$ along $OE$ and it is subjected to a force of $4 	ext{ N}$ in a direction perpendicular to $OE$. The distance of the body from $O$ after $4 	ext{ seconds}$ will be:

Accepted Answer

1. **Analyze the Motion:** The body has an initial velocity along one direction ($x$-axis, along $OE$) and is subjected to a constant force in the perpendicular direction ($y$-axis). This results in a 2D motion with constant velocity in the $x$-direction and constant acceleration in the $y$-direction.

2. **Calculate Acceleration:** Using Newton's Second Law ($F = ma$):
   $$ a_y = \frac{F}{m} = \frac{4 	ext{ N}}{2 	ext{ kg}} = 2 	ext{ m/s}^2 $$
   The acceleration along $OE$ ($a_x$) is zero since there is no force component in that direction [Source 35].

3. **Calculate Displacements:**
   - **Along OE ($x$-axis):** Uniform velocity motion.
     $$ x = u_x t + \frac{1}{2}a_x t^2 $$
     $$ x = 3 	ext{ m/s} 	imes 4 	ext{ s} + 0 = 12 	ext{ m} $$
   - **Perpendicular to OE ($y$-axis):** Uniformly accelerated motion (initial velocity $u_y = 0$).
     $$ y = u_y t + \frac{1}{2}a_y t^2 $$
     $$ y = 0 + \frac{1}{2} 	imes 2 	ext{ m/s}^2 	imes (4 	ext{ s})^2 = 16 	ext{ m} $$
     [Source 30]

4. **Calculate Resultant Distance:** The distance from the origin $O$ is the magnitude of the resultant displacement vector.
   $$ s = \sqrt{x^2 + y^2} $$
   $$ s = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 	ext{ m} $$

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