A man is 45 m behind the bus when the bus starts acceleration from rest with acceleration `2.5(m)/(s^2)`. With what minimum velocity should man start running to catch the bus?

Options

A. 12 m/s
B. 14 m/s
C. 15 m/s
D. 16 m/s

Explanation

To solve the problem of how fast the man should run to catch the bus, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Understand the situation
The bus starts from rest and accelerates at \(2.5 \, \text{m/s}^2\). The man is initially \(45 \, \text{m}\) behind the bus. We need to find the minimum velocity with which the man should start running to catch the bus.

### Step 2: Write down the equations of motion
For the bus, which starts from rest and accelerates:
- Initial velocity \(u_b = 0\)
- Acceleration \(a_b = 2.5 \, \text{m/s}^2\)
- Distance covered by the bus in time \(t\) is given by:
\[ s_b = u_b t + \frac{1}{2} a_b t^2 = 0 + \frac{1}{2} (2.5) t^2 = 1.25 t^2 \]

For the man, who starts running with an initial velocity \(u_m\) and accelerates at \(0 \, \text{m/s}^2\) (constant speed):
- Distance covered by the man in time \(t\) is:
\[ s_m = u_m t \]

### Step 3: Set up the equation to catch the bus
The man needs to cover the initial \(45 \, \text{m}\) gap plus the distance the bus travels in the same time \(t\):
\[ s_m = s_b + 45 \]
Substituting the equations from above:
\[ u_m t = 1.25 t^2 + 45 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 1.25 t^2 - u_m t + 45 = 0 \]
This is a quadratic equation in \(t\).

### Step 5: Use the discriminant for minimum velocity
For the man to catch the bus, the quadratic equation must have real solutions, which means the discriminant must be greater than or equal to zero:
\[ D = b^2 - 4ac \]
Here, \(a = 1.25\), \(b = -u_m\), and \(c = 45\):
\[ D = (-u_m)^2 - 4(1.25)(45) \geq 0 \]
\[ u_m^2 - 225 \geq 0 \]
\[ u_m^2 \geq 225 \]

### Step 6: Solve for \(u_m\)
Taking the square root:
\[ u_m \geq 15 \, \text{m/s} \]

### Conclusion
The minimum velocity with which the man should start running to catch the bus is \(15 \, \text{m/s}\).

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