A man runs at a speed of `4m//s` to overtake a standing bus. When he is `6m` beind the door at `t=0`, the bus mover forward and continuous with a constant acceleration of `1.2m//s^(2)`. The man reaches the door in time `t`. Then

Options

A. `sqrt(3)/2`
B. `sqrt(3)/4`
C. `sqrt(3)`
D. `3/8`

Explanation

To solve the problem step-by-step, let's break it down into manageable parts.

### Step 1: Understand the problem
We have a man running at a speed of \(4 \, \text{m/s}\) and a bus that starts moving from rest with a constant acceleration of \(1.2 \, \text{m/s}^2\). The man is initially \(6 \, \text{m}\) behind the bus door when \(t = 0\).

### Step 2: Write the equations for the distances
1. **Distance traveled by the bus**:
The bus starts from rest and moves with constant acceleration. The distance \(d_b\) traveled by the bus in time \(t\) can be calculated using the equation:
\[ d_b = \frac{1}{2} a t^2 \]
where \(a = 1.2 \, \text{m/s}^2\).
Thus,
\[ d_b = \frac{1}{2} \times 1.2 \times t^2 = 0.6 t^2 \]

2. **Distance traveled by the man**:
The man runs at a constant speed of \(4 \, \text{m/s}\). The distance \(d_m\) he covers in time \(t\) is given by:
\[ d_m = 4t \]

### Step 3: Set up the equation
At the moment the man reaches the bus door, the distance he has covered must equal the distance between him and the bus door plus the distance the bus has traveled. Therefore, we can write:
\[ d_m = 6 + d_b \]
Substituting the expressions for \(d_m\) and \(d_b\):
\[ 4t = 6 + 0.6t^2 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 0.6t^2 - 4t + 6 = 0 \]

### Step 5: Solve the quadratic equation
To solve the quadratic equation \(0.6t^2 - 4t + 6 = 0\), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 0.6\), \(b = -4\), and \(c = 6\).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \times 0.6 \times 6 = 16 - 14.4 = 1.6 \]

Now substituting into the quadratic formula:
\[ t = \frac{4 \pm \sqrt{1.6}}{2 \times 0.6} \]
Calculating \(\sqrt{1.6} \approx 1.2649\):
\[ t = \frac{4 \pm 1.2649}{1.2} \]
Calculating the two possible values for \(t\):
1. \(t_1 = \frac{4 + 1.2649}{1.2} \approx 4.25\)
2. \(t_2 = \frac{4 - 1.2649}{1.2} \approx 2.29\)

### Step 6: Conclusion
The time \(t\) at which the man reaches the bus door can be either \(4.25\) seconds or \(2.29\) seconds depending on the context of the problem.