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**Hint:**The distance travelled by the student must be 50m more than the distance travelled by bus at the same time, so as to catch the bus. The equations of motion for both student and bus have to be of the same time.

**Complete step by step answer:**In order to catch the bus, the student has to travel the distance travelled by bus in that time as well as another 50m because it is that much distance away from the bus initially.

So, we write the equation of motion for the student as

$x + 50 = ut$ _____________1

where x is the distance travelled by bus

u is the velocity of the student

t is the time in which he catches the bus

The equation of motion for the bus is,

$

x = vt + \dfrac{1}{2}a{t^2} \\

x = \dfrac{1}{2}a{t^2} \\

$ ___________2

where x is the distance travelled by bus

v is the initial velocity of bus (v=0)

a is the acceleration of the bus

t is the time in which the student catches the bus

Substituting the value of equation 2 in equation 1, we get

\[

\dfrac{1}{2}a{t^2} + 50 = ut \\

a{t^2} - 2ut + 100 = 0 \\

\]

\[{t^2} - 2ut + 100 = 0\] (because a=1m/s2)

The above equation must have real roots because time has to be positive and real, therefore, the discriminant (D) of the equation must be greater than or equal to zero.

$

D \geqslant 0 \\

{\left( { - 2u} \right)^2} - 4 \times 1 \times 100 \geqslant 0 \\

{u^2} \geqslant 100 \\

\therefore u \geqslant 10m/s \\

$

Therefore, the correct option is D) 10m/s

**Note:**The extra fifty meter distance has to be kept in notice.

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