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: \[ 30 { m}. \]

Answer 2

To solve this problem, we will determine the displacement of the body interacting with forces in two orthogonal directions, resulting in motion described by vector addition.
The body initially moves south with a velocity: \( v_s = 8 \text{ m/s} \) and has a mass \( m = 1.5 \text{ kg} \). A force \( F = 6 \text{ N} \) is applied east for a duration \( t = 3 \text{ s} \).
The acceleration \( a \) due to the force is calculated using Newton’s second law: \( F = ma \),
\( a = \frac{F}{m} = \frac{6}{1.5} = 4 \text{ m/s}^2 \).
Now, we will calculate the components of displacement.

Southward Displacement \( (s_s) \):	\( s_s = v_s \cdot t = 8 \cdot 3 = 24 \text{ m} \)
Eastward Displacement \( (s_e) \):	Using the equation \( s = ut + \frac{1}{2}at^2 \) where initial velocity \( u = 0\): \( s_e = 0 \cdot 3 + \frac{1}{2}(4)(3)^2 = \frac{1}{2} \cdot 4 \cdot 9 = 18 \text{ m} \)

Using the Pythagorean theorem, total displacement \( S \) is:
\( S = \sqrt{s_s^2 + s_e^2} = \sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = 30 \text{ m} \).
Hence, the displacement of the body after 3 seconds is \( 30 \) meters.