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 of finding the magnitude of the velocity of a motorboat after its engine is cut off, we start with the given deceleration equation:

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

where \( k \) is a constant and \( v_0 \) is the initial velocity at the time of cut-off.

### Step 1: Separate Variables
We can rearrange the equation to separate the variables \( v \) and \( t \):

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

### Step 2: Integrate Both Sides
Next, we 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 \( t \):

\[ \int_{v_0}^{v} \frac{dv}{v^3} = -k \int_{0}^{t} dt \]

### Step 3: Calculate the Integrals
Calculating the left side:

\[ \int \frac{dv}{v^3} = -\frac{1}{2v^2} \]

So we have:

\[ \left[-\frac{1}{2v^2}\right]_{v_0}^{v} = -kt \]

This gives us:

\[ -\frac{1}{2v^2} + \frac{1}{2v_0^2} = -kt \]

### Step 4: Rearranging the Equation
Rearranging the equation, we can multiply through by -1:

\[ \frac{1}{2v^2} - \frac{1}{2v_0^2} = kt \]

Multiplying through by 2 gives:

\[ \frac{1}{v^2} - \frac{1}{v_0^2} = 2kt \]

### Step 5: Solve for \( v^2 \)
Now, we can rearrange this to find \( v^2 \):

\[ \frac{1}{v^2} = 2kt + \frac{1}{v_0^2} \]

Taking the reciprocal gives:

\[ v^2 = \frac{1}{2kt + \frac{1}{v_0^2}} \]

### Step 6: Final Expression for \( v \)
To express \( v \) in terms of \( v_0 \) and \( t \):

\[ v = \frac{v_0}{\sqrt{1 + 2ktv_0^2}} \]

### Conclusion
Thus, the magnitude of the velocity at time \( t \) after the cut-off is:

\[ v = \frac{v_0}{\sqrt{1 + 2ktv_0^2}} \]