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**.