A bus of mass 1000 kg has an engine which produces a constant power of 50 kW. If the resistance to motion, assumed constant is 1000 N. The maximum speed at which the bus can travel on level road and the acceleration when it is travelling at 25 m/s, will respectively be -

Answer

To solve the problem step by step, we need to find two things: the maximum speed of the bus and the acceleration when it is traveling at 25 m/s.

### Given Data:
- Mass of the bus (m) = 1000 kg
- Power of the engine (P) = 50 kW = 50,000 W
- Resistance to motion (F_resistance) = 1000 N
- Initial speed (v_i) = 25 m/s

### Step 1: Calculate the net force acting on the bus.
The net force (F_net) acting on the bus can be calculated using the formula:
\[ F_{net} = P/v - F_{resistance} \]
Where \( v \) is the speed of the bus.

### Step 2: Find the maximum speed of the bus.
At maximum speed, the power produced by the engine is equal to the power required to overcome the resistance. Thus, we can set up the equation:
\[ P = F_{resistance} \cdot v_{max} \]
Rearranging gives:
\[ v_{max} = \frac{P}{F_{resistance}} \]

Substituting the values:
\[ v_{max} = \frac{50,000 \, W}{1000 \, N} = 50 \, m/s \]

### Step 3: Calculate the acceleration when the bus is traveling at 25 m/s.
To find the acceleration (a) when the bus is traveling at 25 m/s, we can use Newton's second law:
\[ F_{net} = m \cdot a \]
Where:
\[ F_{net} = P/v - F_{resistance} \]

First, we calculate the force produced by the engine at 25 m/s:
\[ F_{engine} = \frac{P}{v} = \frac{50,000 \, W}{25 \, m/s} = 2000 \, N \]

Now, we can find the net force:
\[ F_{net} = F_{engine} - F_{resistance} = 2000 \, N - 1000 \, N = 1000 \, N \]

Now, using Newton's second law to find acceleration:
\[ a = \frac{F_{net}}{m} = \frac{1000 \, N}{1000 \, kg} = 1 \, m/s^2 \]

### Final Answers:
- Maximum speed of the bus, \( v_{max} = 50 \, m/s \)
- Acceleration at 25 m/s, \( a = 1 \, m/s^2 \)