A man is 25 m behind a bus, when bus starts accelerating at `2 ms^(-2)` and man starts moving with constant velocity of `10 ms^(-1)`. Time taken by him to board the bus is

Options

A. 2 s
B. 3 s
C. 4 s
D. 5 s

Explanation

To solve the problem step by step, we need to analyze the motion of both the man and the bus.

### Step 1: Define the Displacements
Let:
- \( s_m \) be the displacement of the man.
- \( s_b \) be the displacement of the bus.

The man starts 25 m behind the bus, so we can express the displacement of the man as:
\[ s_m = s_b + 25 \]

### Step 2: Write the Equations of Motion
1. **For the man** (moving with constant velocity):
\[ s_m = u_m t = 10t \]
where \( u_m = 10 \, \text{m/s} \) (the man's velocity) and \( t \) is the time in seconds.

2. **For the bus** (starting from rest and accelerating):
\[ s_b = u_b t + \frac{1}{2} a_b t^2 = 0 + \frac{1}{2} (2) t^2 = t^2 \]
where \( u_b = 0 \, \text{m/s} \) (the bus's initial velocity) and \( a_b = 2 \, \text{m/s}^2 \) (the bus's acceleration).

### Step 3: Set Up the Equation
From the previous steps, we have:
\[ 10t = t^2 + 25 \]

### Step 4: Rearrange the Equation
Rearranging gives us:
\[ t^2 - 10t + 25 = 0 \]

### Step 5: Solve the Quadratic Equation
This can be factored as:
\[ (t - 5)^2 = 0 \]
Thus, we find:
\[ t - 5 = 0 \quad \Rightarrow \quad t = 5 \, \text{seconds} \]

### Conclusion
The time taken by the man to board the bus is \( t = 5 \) seconds.

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