A car of mass `m` is driven with acceleration `a` along a straight level road against a constant external resistive force R. When the velocity of the car V, the rate at which the engine of the car is doing work will be

Options

A. RV
B. maV
C. (R+ma)V
D. (ma-R)V

Explanation

To solve the problem, we need to determine the rate at which the engine of the car is doing work when it is moving with a certain velocity \( V \), given that it has a mass \( m \), an acceleration \( a \), and is facing a resistive force \( R \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Car:**
- The car is driven by an engine force \( F \) in the direction of motion.
- There is a resistive force \( R \) acting in the opposite direction.

2. **Apply Newton's Second Law:**
- According to Newton's second law, the net force acting on the car is equal to the mass of the car multiplied by its acceleration:
\[ F_{\text{net}} = ma \]
- The net force can also be expressed as:
\[ F_{\text{net}} = F - R \]
- Setting these two expressions for net force equal gives:
\[ F - R = ma \]

3. **Solve for the Engine Force \( F \):**
- Rearranging the equation gives:
\[ F = ma + R \]

4. **Calculate the Power Delivered by the Engine:**
- The power \( P \) delivered by the engine is given by the product of the force exerted by the engine and the velocity of the car:
\[ P = F \cdot V \]
- Substituting the expression for \( F \) from step 3 into this equation gives:
\[ P = (ma + R) \cdot V \]

5. **Final Expression for Power:**
- Thus, the rate at which the engine of the car is doing work is:
\[ P = (ma + R)V \]

### Conclusion:
The rate at which the engine of the car is doing work is given by:
\[ P = (ma + R)V \]