A car of mas 1250 kg experiences a resistance of 750 n when it moves at `30ms^(-1)`. If the engine can develop 30 KW at this speed, the maximum acceleration that the engine can produce is

Options

A. `0.8ms^(-2)`
B. `0.2ms^(-2)`
C. `0.4ms^(-2)`
D. `0.5ms^(-2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Mass of the car, \( m = 1250 \, \text{kg} \)
- Resistance force, \( f = 750 \, \text{N} \)
- Speed of the car, \( v = 30 \, \text{m/s} \)
- Power developed by the engine, \( P = 30 \, \text{kW} = 30 \times 10^3 \, \text{W} \)

### Step 2: Calculate the force exerted by the engine
The power developed by the engine can be expressed in terms of force and velocity:
\[ P = F \cdot v \]
From this, we can rearrange to find the force \( F \):
\[ F = \frac{P}{v} \]
Substituting the values:
\[ F = \frac{30 \times 10^3 \, \text{W}}{30 \, \text{m/s}} = 1000 \, \text{N} \]

### Step 3: 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 \( a \):
\[ F_{\text{net}} = m \cdot a \]
The net force is the difference between the force exerted by the engine and the resistance force:
\[ F_{\text{net}} = F - f \]
Substituting the values we found:
\[ F_{\text{net}} = 1000 \, \text{N} - 750 \, \text{N} = 250 \, \text{N} \]

### Step 4: Calculate the acceleration
Now we can substitute \( F_{\text{net}} \) into Newton's second law:
\[ m \cdot a = F_{\text{net}} \]
Thus,
\[ a = \frac{F_{\text{net}}}{m} = \frac{250 \, \text{N}}{1250 \, \text{kg}} = 0.2 \, \text{m/s}^2 \]

### Final Answer
The maximum acceleration that the engine can produce is:
\[ \boxed{0.2 \, \text{m/s}^2} \]

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