A bus increases its speed from 36 km/h to 72 km/h in 10 seconds. Its acceleration is :

Options

A. `5 m//s^(2)`
B. `2m//s^(2)`
C. `3.6 m//s^(2)`
D. `1 m//s^(2)`

Answer

To find the acceleration of the bus that increases its speed from 36 km/h to 72 km/h in 10 seconds, we can follow these steps:

### Step 1: Identify the initial and final velocities
- Initial velocity (U) = 36 km/h
- Final velocity (V) = 72 km/h

### Step 2: Convert velocities from km/h to m/s
To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \).

- Convert U:
\[ U = 36 \text{ km/h} \times \frac{5}{18} = 10 \text{ m/s} \]

- Convert V:
\[ V = 72 \text{ km/h} \times \frac{5}{18} = 20 \text{ m/s} \]

### Step 3: Use the formula for acceleration
The formula for acceleration (a) is given by:
\[ a = \frac{V - U}{t} \]
where:
- \( V \) = final velocity
- \( U \) = initial velocity
- \( t \) = time taken

### Step 4: Substitute the values into the formula
Substituting the values we have:
- \( V = 20 \text{ m/s} \)
- \( U = 10 \text{ m/s} \)
- \( t = 10 \text{ seconds} \)

So,
\[ a = \frac{20 \text{ m/s} - 10 \text{ m/s}}{10 \text{ s}} = \frac{10 \text{ m/s}}{10 \text{ s}} = 1 \text{ m/s}^2 \]

### Step 5: Conclusion
The acceleration of the bus is \( 1 \text{ m/s}^2 \).

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