A block is thrown with a velocity of `2m//s^(-1)` (relative to ground) on a belt, which is moving with velocity `4ms^(-1)` in opposite direction of the initial velocity of block. If the block stops slipping on the belt after `4sec` of the throwing then choose the correct statements `(s)`

Options

A. Displacement with respect to ground is zero after 2.66 s and magnitude of displacement with respect to ground is 12 m after 4s.
B. Magnitude of displacement with respect to ground in 4s is 4 m.
C. Magnitude of displacement with respect to belt in 4s is 12 m.
D. Displacement with respect to ground is zero in 8/3 s.

Answer

To solve the problem step by step, we will analyze the motion of the block and the belt separately and then combine the results to determine the correct statements.

### Step 1: Understand the Initial Conditions
- The block is thrown with a velocity of \( v_b = 2 \, \text{m/s} \) (relative to the ground).
- The belt is moving with a velocity of \( v_{belt} = 4 \, \text{m/s} \) in the opposite direction to the block's initial velocity.

### Step 2: Determine the Initial Relative Velocity
Since the block is thrown in the opposite direction to the belt's motion, the initial velocity of the block relative to the belt can be calculated as:
\[ v_{relative} = v_b - v_{belt} = 2 \, \text{m/s} - (-4 \, \text{m/s}) = 2 + 4 = 6 \, \text{m/s} \]

### Step 3: Calculate the Acceleration of the Block
The block stops slipping after \( t = 4 \, \text{s} \). We can use the formula for final velocity:
\[ v = u + at \]
Where:
- \( v = 0 \) (the block stops slipping),
- \( u = 6 \, \text{m/s} \) (initial relative velocity),
- \( t = 4 \, \text{s} \).

Rearranging gives:
\[ 0 = 6 + a(4) \]
Solving for \( a \):
\[ a = -\frac{6}{4} = -1.5 \, \text{m/s}^2 \]

### Step 4: Calculate the Displacement of the Block
Using the equation of motion:
\[ x = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ x = 6 \times 4 + \frac{1}{2} \times (-1.5) \times (4^2) \]
Calculating:
\[ x = 24 - \frac{1}{2} \times 1.5 \times 16 = 24 - 12 = 12 \, \text{m} \]

### Step 5: Calculate the Displacement of the Belt
The displacement of the belt during the same time can be calculated as:
\[ x_{belt} = v_{belt} \times t \]
Substituting the values:
\[ x_{belt} = 4 \times 4 = 16 \, \text{m} \]

### Step 6: Analyze the Results
- The block moves \( 12 \, \text{m} \) relative to the ground.
- The belt moves \( 16 \, \text{m} \) relative to the ground.

### Conclusion
The block stops slipping after \( 4 \, \text{s} \) and the distances covered are:
- Block: \( 12 \, \text{m} \)
- Belt: \( 16 \, \text{m} \)

Now, we can choose the correct statements based on these calculations.