A student is standing at a distance of `50` metres from a bus. As soon as the bus begins its motion (starts moving away from student) with an acceleration of `1 ms^-2`, the student starts running towards the bus with a uniform velocity `u` . Assuming the motion to be along a straight road. The minimum value of `u`, so that the student is able to catch the bus is :

Options

A. `5 ms^-1`
B. `8 ms^-1`
C. `10 m//s^-1`
D. `12 m//s^-1`

Explanation

To solve the problem, we need to analyze the motion of both the bus and the student. The bus starts moving away from the student with an acceleration of \(1 \, \text{m/s}^2\), while the student runs towards the bus with a uniform velocity \(u\). We want to find the minimum value of \(u\) such that the student can catch the bus.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The initial distance between the student and the bus is \(s = 50 \, \text{m}\).
- The bus starts from rest, so its initial velocity \(v_0 = 0\).
- The acceleration of the bus is \(a = 1 \, \text{m/s}^2\).
- The student runs towards the bus with a uniform velocity \(u\).

2. **Equation of Motion for the Bus:**
The distance covered by the bus after time \(t\) can be described by the equation of motion:
\[ s_{\text{bus}} = v_0 t + \frac{1}{2} a t^2 \]
Since \(v_0 = 0\), this simplifies to:
\[ s_{\text{bus}} = \frac{1}{2} a t^2 = \frac{1}{2} (1) t^2 = \frac{1}{2} t^2 \]

3. **Distance Covered by the Student:**
The distance covered by the student in the same time \(t\) is given by:
\[ s_{\text{student}} = u t \]

4. **Relative Distance:**
The student needs to cover the initial distance of \(50 \, \text{m}\) plus the distance the bus has moved away from the starting point. Therefore, the equation becomes:
\[ s_{\text{student}} = s + s_{\text{bus}} \]
Substituting the expressions we derived:
\[ u t = 50 + \frac{1}{2} t^2 \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{1}{2} t^2 - u t + 50 = 0 \]

6. **Using the Quadratic Formula:**
This is a quadratic equation in \(t\). For the student to catch the bus, this equation must have real solutions, which means the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \(a = \frac{1}{2}\), \(b = -u\), and \(c = 50\). Thus:
\[ D = (-u)^2 - 4 \left(\frac{1}{2}\right)(50) \geq 0 \]
Simplifying gives:
\[ u^2 - 100 \geq 0 \]

7. **Solving for \(u\):**
This implies:
\[ u^2 \geq 100 \]
Taking the square root of both sides, we find:
\[ u \geq 10 \, \text{m/s} \]

### Conclusion:
The minimum value of \(u\) so that the student is able to catch the bus is \(10 \, \text{m/s}\).

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