Question

A body of mass 1.5 kg is moving towards south with a uniform velocity of \( 8 { ms}^{-1} \). A force of \( 6 \) N is applied to the body towards east. The displacement of the body 3 seconds after the application of the force is:

• A. \( 24 \) m
• B. \( 30 \) m
• C. \( 18 \) m
• D. \( 42 \) m

Hint:

When forces act in perpendicular directions, use kinematic equations separately for each direction and apply the Pythagorean theorem for net displacement.

Answer 1

The Correct Option is B

Step 1: Understanding the motion in two perpendicular directions
- Initial velocity towards south: \( v_y = 8 { ms}^{-1} \).
- A force of \( 6 \) N is applied towards east.
- Mass of the body: \( m = 1.5 \) kg.
- Time of action: \( t = 3 \) s.
Using Newton's second law, acceleration in the eastward direction is: \[ a_x = \frac{F}{m} = \frac{6}{1.5} = 4 { ms}^{-2}. \] Step 2: Computing displacement in each direction
Since the body is initially moving south, its displacement in the south direction in 3 s is: \[ S_y = v_y t = 8 \times 3 = 24 { m}. \] For eastward motion (starting from rest): \[ S_x = \frac{1}{2} a_x t^2. \] \[ S_x = \frac{1}{2} \times 4 \times (3)^2. \] \[ S_x = \frac{1}{2} \times 4 \times 9 = 18 { m}. \] Step 3: Finding resultant displacement
Since the displacements \( S_x \) and \( S_y \) are perpendicular, the net displacement is given by: \[ S = \sqrt{S_x^2 + S_y^2}. \] \[ S = \sqrt{(18)^2 + (24)^2}. \] \[ S = \sqrt{324 + 576} = \sqrt{900} = 30 { m}. \] Step 4: Conclusion
Thus, the total displacement of the body after 3 seconds is: \[ 30 { m}. \]