The retardation experienced by a moving motor bike after its engine is cut-off , at the instant (t) is given by, `a =-k v^4` , where (k) is a constant. If ` v_0` is the magnitude of velocity at the cut-off , the magnitude of velocity at time (t) after the cut-off is .

Answer

To solve the problem, we need to find the magnitude of the velocity \( v \) of the motorbike at time \( t \) after the engine is cut off, given the retardation \( a = -k v^4 \).

### Step-by-Step Solution:

1. **Understand the given information**:
- Retardation (deceleration) is given by the equation:
\[ a = -k v^4 \]
- The initial velocity at the moment the engine is cut off is \( v_0 \).

2. **Relate acceleration to velocity**:
- We know that acceleration can also be expressed as:
\[ a = \frac{dv}{dt} \]
- Therefore, we can write:
\[ \frac{dv}{dt} = -k v^4 \]

3. **Separate variables**:
- Rearranging the equation gives:
\[ \frac{dv}{v^4} = -k dt \]

4. **Integrate both sides**:
- We will integrate the left side from \( v_0 \) to \( v \) and the right side from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v^4} = -k \int_{0}^{t} dt \]

5. **Calculate the integrals**:
- The integral on the left side:
\[ \int \frac{dv}{v^4} = -\frac{1}{3 v^3} \]
- Thus, we have:
\[ \left[-\frac{1}{3 v^3}\right]_{v_0}^{v} = -kt \]
- This leads to:
\[ -\frac{1}{3 v^3} + \frac{1}{3 v_0^3} = -kt \]

6. **Rearranging the equation**:
- Multiply through by -3:
\[ \frac{1}{v^3} - \frac{1}{v_0^3} = 3kt \]
- Rearranging gives:
\[ \frac{1}{v^3} = 3kt + \frac{1}{v_0^3} \]

7. **Find \( v \)**:
- Taking the reciprocal:
\[ v^3 = \frac{1}{3kt + \frac{1}{v_0^3}} \]
- Therefore, we can express \( v \) as:
\[ v = \left( \frac{1}{3kt + \frac{1}{v_0^3}} \right)^{\frac{1}{3}} \]

### Final Result:
The magnitude of the velocity at time \( t \) after the cut-off is:
\[ v = \frac{v_0}{(1 + 3k v_0^3 t)^{\frac{1}{3}}} \]