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 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**.