The deceleration exerienced by a moving motor blat, after its engine is cut-off is given by `dv//dt=-kv^(3)`, where `k` is constant. If `v_(0)` is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time `t` after the cut-off is.

Options

A. `v_(0)//2`
B. `v`
C. `v_(0)e^(-kt)`
D. `(v_(0))/(sqrt(2v_(0)^(2)kt+1))`

Explanation

To solve the problem, we need to find the velocity \( v \) of the motorboat at a time \( t \) after the engine is cut off, given the deceleration equation:

\[ \frac{dv}{dt} = -k v^3 \]

where \( k \) is a constant and \( v_0 \) is the initial velocity at the moment of cut-off (i.e., at \( t = 0 \)).

### Step-by-Step Solution:

1. **Rearranging the Equation:**
We start with the given equation:
\[ \frac{dv}{dt} = -k v^3 \]
Rearranging gives:
\[ \frac{dv}{v^3} = -k dt \]

2. **Integrating Both Sides:**
We will integrate both sides. The left side will be integrated with respect to \( v \) from \( v_0 \) to \( v \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v^3} = -k \int_{0}^{t} dt \]

3. **Calculating the Integrals:**
The integral on the left side:
\[ \int \frac{dv}{v^3} = -\frac{1}{2v^2} \]
Thus, we have:
\[ -\frac{1}{2v^2} \bigg|_{v_0}^{v} = -kt \]
This gives:
\[ -\frac{1}{2v^2} + \frac{1}{2v_0^2} = -kt \]

4. **Rearranging the Equation:**
Rearranging the above equation:
\[ \frac{1}{2v_0^2} - \frac{1}{2v^2} = kt \]
Multiplying through by 2:
\[ \frac{1}{v_0^2} - \frac{1}{v^2} = 2kt \]

5. **Finding a Common Denominator:**
Rearranging gives:
\[ \frac{1}{v^2} = \frac{1}{v_0^2} - 2kt \]

6. **Taking the Reciprocal:**
Taking the reciprocal of both sides gives:
\[ v^2 = \frac{1}{\frac{1}{v_0^2} - 2kt} \]

7. **Final Expression for Velocity:**
Thus, we can express \( v \) as:
\[ v = \frac{v_0}{\sqrt{1 - 2ktv_0^2}} \]

### Final Answer:
The magnitude of the velocity at time \( t \) after the cut-off is given by:
\[ v = \frac{v_0}{\sqrt{1 + 2ktv_0^2}} \]