A man is at a distance of 6 m from a bus. The bus begins to move with a constant acceleration of `3 ms^(−2)`. In order to catch the bus, the minimum speed with which the man should run towards the bus is

Options

A. `2 ms^(-1)`
B. `4 ms^(-1)`
C. `6 ms^(-1)`
D. `8 ms^(-1)`

Answer

To solve the problem, we need to find the minimum speed \( V \) with which the man should run towards the bus in order to catch it. Let's break down the steps:

### Step 1: Understand the scenario
The man is initially 6 meters away from the bus, which starts moving with a constant acceleration of \( a = 3 \, \text{m/s}^2 \). The bus starts from rest, so its initial velocity \( u = 0 \).

### Step 2: Use the equations of motion
We can use the equations of motion to relate the distance traveled by the bus to the speed of the man. The equation we will use is:

\[ V^2 = u^2 + 2as \]

Where:
- \( V \) is the final velocity of the bus,
- \( u \) is the initial velocity of the bus (which is 0),
- \( a \) is the acceleration of the bus,
- \( s \) is the distance traveled by the bus.

### Step 3: Substitute the known values
Since the bus starts from rest, \( u = 0 \). The acceleration \( a = 3 \, \text{m/s}^2 \) and the distance \( s = 6 \, \text{m} \) (the distance between the man and the bus). Plugging these values into the equation gives:

\[ V^2 = 0 + 2 \times 3 \times 6 \]

### Step 4: Calculate \( V^2 \)
Now, we calculate \( V^2 \):

\[ V^2 = 2 \times 3 \times 6 = 36 \]

### Step 5: Find \( V \)
To find \( V \), we take the square root of \( V^2 \):

\[ V = \sqrt{36} = 6 \, \text{m/s} \]

### Conclusion
Thus, the minimum speed with which the man should run towards the bus to catch it is \( 6 \, \text{m/s} \).