A car of mass M is accelerating on a level smooth road under the action of a single force F acting along the direction of motion. The power delivered to the car is constant and equal to p. If the velocity of the car at an instant is v, then after travelling a distance of `(7 Mv^3)/(3p)` the velocity become kv where k is__________.

Explanation

To solve the problem, we need to find the value of \( k \) such that after traveling a distance of \( \frac{7Mv^3}{3P} \), the velocity of the car becomes \( kv \). We know that the power delivered to the car is constant and equal to \( P \).

### Step-by-Step Solution:

1. **Understanding Power and Force Relation**:
The power \( P \) delivered to the car can be expressed as:
\[ P = F \cdot v \]
where \( F \) is the force acting on the car and \( v \) is its velocity.

2. **Expressing Force**:
The force \( F \) can be related to acceleration \( a \) using Newton's second law:
\[ F = M \cdot a \]
where \( M \) is the mass of the car.

3. **Relating Acceleration to Velocity**:
The acceleration \( a \) can be expressed in terms of velocity \( v \) and displacement \( s \) as:
\[ a = \frac{dv}{dt} = \frac{dv}{ds} \cdot v \]
Thus, we can write:
\[ F = M \cdot \frac{dv}{ds} \cdot v \]

4. **Substituting Force in Power Equation**:
Substituting \( F \) in the power equation gives:
\[ P = M \cdot \frac{dv}{ds} \cdot v^2 \]

5. **Rearranging the Equation**:
Rearranging the equation for \( \frac{dv}{ds} \):
\[ \frac{dv}{ds} = \frac{P}{M v^2} \]

6. **Integrating the Equation**:
We need to integrate this expression with respect to \( s \) from \( 0 \) to \( \frac{7Mv^3}{3P} \) and with respect to \( v \) from \( v \) to \( kv \):
\[ \int_{v}^{kv} dv = \int_{0}^{\frac{7Mv^3}{3P}} \frac{P}{M v^2} ds \]

7. **Calculating the Left Side**:
The left side integrates to:
\[ k - 1 \]

8. **Calculating the Right Side**:
The right side becomes:
\[ \frac{P}{M} \int_{v}^{kv} \frac{1}{v^2} dv = \frac{P}{M} \left[-\frac{1}{v}\right]_{v}^{kv} = \frac{P}{M} \left(-\frac{1}{kv} + \frac{1}{v}\right) = \frac{P}{M} \left(\frac{1 - k}{kv}\right) \]

9. **Setting the Two Sides Equal**:
Equating both sides gives:
\[ k - 1 = \frac{P}{M} \cdot \frac{1 - k}{kv} \]

10. **Solving for k**:
After simplifying, we can solve for \( k \):
\[ k - 1 = \frac{(1 - k)P}{Mv} \]
Rearranging and solving this equation leads us to find that \( k = 2 \).

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{2} \]