If acceleration gradient of a moving body along positive x-axis is `-4s^(-2)` and body starts its motion from rest from `x = 5m`, having maximum acceleration (magnitude) then velocity of moving body at `x = -3m` is

Options

A. `6ms^(-1)`
B. `4 ms^(-1)`
C. `8 ms^(-1)`
D. Can't reach at `x = -3m`

Explanation

To solve the problem, we need to find the velocity of a moving body at \( x = -3 \, \text{m} \) given that the acceleration gradient is \( -4s^{-2} \) and the body starts from rest at \( x = 5 \, \text{m} \).

### Step-by-Step Solution:

1. **Understand the acceleration gradient**:
The acceleration gradient is given as \( \frac{dA}{dx} = -4 \). This means that the acceleration \( A \) can be expressed as:
\[ A = -4x + C \]
where \( C \) is a constant.

2. **Integrate to find velocity**:
We know that acceleration \( A \) can also be expressed as:
\[ A = v \frac{dv}{dx} \]
Setting the two expressions for acceleration equal gives:
\[ v \frac{dv}{dx} = -4x + C \]

3. **Rearranging and integrating**:
Rearranging gives:
\[ v \, dv = (-4x + C) \, dx \]
Integrating both sides:
\[ \int v \, dv = \int (-4x + C) \, dx \]
This results in:
\[ \frac{v^2}{2} = -2x^2 + Cx + k \]
where \( k \) is a constant of integration.

4. **Using initial conditions**:
The body starts from rest at \( x = 5 \, \text{m} \). Therefore, when \( x = 5 \), \( v = 0 \):
\[ 0 = -2(5^2) + C(5) + k \]
Simplifying gives:
\[ 0 = -50 + 5C + k \]
Rearranging gives:
\[ k = 50 - 5C \]

5. **Substituting back**:
Substituting \( k \) back into the equation for \( v^2 \):
\[ v^2 = -4x^2 + Cx + (50 - 5C) \]

6. **Finding \( C \)**:
To find \( C \), we need to determine the maximum acceleration. The maximum acceleration occurs when \( \frac{dA}{dx} = 0 \):
\[ -4 = 0 \implies \text{not applicable, we need to find the maximum value of } A. \]
The maximum occurs at \( x = 0 \):
\[ A = C \]

7. **Substituting \( x = -3 \)**:
Now we need to find \( v^2 \) at \( x = -3 \):
\[ v^2 = -4(-3)^2 + C(-3) + (50 - 5C) \]
This simplifies to:
\[ v^2 = -36 - 3C + 50 - 5C = 14 - 8C \]

8. **Finding the condition for \( v^2 \)**:
For \( v^2 \) to be non-negative:
\[ 14 - 8C \geq 0 \implies C \leq \frac{14}{8} = 1.75 \]

9. **Conclusion**:
Since the body starts from rest and the maximum acceleration is at \( x = 0 \), the body cannot reach \( x = -3 \) because it cannot achieve the necessary velocity to do so. Thus, the answer is that the body cannot reach \( x = -3 \, \text{m} \).

### Final Answer:
The velocity of the moving body at \( x = -3 \, \text{m} \) is **not achievable** (the body cannot reach this position).