A student is standing at a distance of 50 metres from the bus. As soon as the bus starts its motion with an acceleration of 1m/s², 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 students is able to catch the bus is

The correct option is 10 m/s.

Let's solve this step-by-step to find the minimum uniform velocity $u$ required by the student to catch the bus.

Step 1: Understand the motion of both objects
Let the initial position of the bus be the origin ($x=0$).
The student is initially standing at a distance of 50 meters behind the bus, so the student's initial position is:
$x_s\left(0\right)=-50\text{m}$

The bus starts from rest with a constant acceleration:
$a=1{\text{m/s}}^2$
Its position at any time $t$ is given by the equation of motion:
$x_b\left(t\right)=x_b\left(0\right)+v_{0}t+\frac{1}{2}a{t}^2$
Since $x_b\left(0\right)=0$ and $v_0=0$:
$x_b\left(t\right)=\frac{1}{2}\left(1\right)t^2=\frac{t^2}{2}$

The student runs towards the bus with a uniform velocity $u$ starting from $x=-50\text{m}$.
The position of the student at any time $t$ is given by:
$x_s\left(t\right)=-50+ut$

Step 2: Condition for the student to catch the bus
The student catches the bus when their positions are equal, i.e., $x_s\left(t\right)=x_b\left(t\right)$:
$-50+ut=\frac{t^2}{2}$

Rearranging this equation into a standard quadratic form in terms of $t$:
$t^2-2ut+100=0$

Step 3: Find the condition for real time $t$
For the student to catch the bus, the time $t$ must be a real value. Therefore, the discriminant of the quadratic equation must be greater than or equal to zero:
$D=B^2-4AC\ge 0$

Substitute $A=1$, $B=-2u$, and $C=100$ into the inequality:
${(-2u)}^2-4\left(1\right)\left(100\right)\ge 0$
$4u^2-400\ge 0$
$u^2\ge 100$
$u\ge 10\text{m/s}$

Conclusion:
The minimum value of velocity $u$ required for the student to catch the bus is 10 m/s.