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

Answer

To solve the problem, we need to determine the minimum speed at which a man must run towards a bus that is accelerating away from him. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The man is initially 6 meters away from the bus, which starts moving with a constant acceleration of \( a = 3 \, \text{m/s}^2 \). We need to find the minimum speed \( v \) at which the man must run to catch the bus.

### Step 2: Set Up the Equations
1. **Distance covered by the bus**: The bus starts from rest and accelerates. The distance \( s_2 \) covered by the bus in time \( t \) is given by the equation:
\[ s_2 = \frac{1}{2} a t^2 \]
Substituting \( a = 3 \, \text{m/s}^2 \):
\[ s_2 = \frac{1}{2} \cdot 3 \cdot t^2 = \frac{3}{2} t^2 \]

2. **Distance covered by the man**: The man runs towards the bus with speed \( v \). The distance \( s_1 \) he covers in time \( t \) is:
\[ s_1 = v t \]

### Step 3: Relate the Distances
To catch the bus, the total distance covered by the man must equal the initial distance plus the distance the bus has moved:
\[ s_1 = 6 + s_2 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ v t = 6 + \frac{3}{2} t^2 \]

### Step 4: Rearrange the Equation
Rearranging the equation gives us:
\[ \frac{3}{2} t^2 - v t + 6 = 0 \]
This is a quadratic equation in \( t \).

### Step 5: Apply the Quadratic Formula
For the quadratic equation \( at^2 + bt + c = 0 \), the solutions for \( t \) are given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = \frac{3}{2} \), \( b = -v \), and \( c = 6 \):
\[ t = \frac{v \pm \sqrt{v^2 - 4 \cdot \frac{3}{2} \cdot 6}}{2 \cdot \frac{3}{2}} \]
This simplifies to:
\[ t = \frac{v \pm \sqrt{v^2 - 36}}{3} \]

### Step 6: Ensure Real Solutions
For \( t \) to be real, the discriminant must be non-negative:
\[ v^2 - 36 \geq 0 \]
This implies:
\[ v^2 \geq 36 \]
Thus:
\[ v \geq 6 \]

### Step 7: Conclusion
The minimum speed \( v \) that the man must run to catch the bus is:
\[ v = 6 \, \text{m/s} \]

Thus, the answer is **6 meters per second**.