Question

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 $1m{s}^{-2}$, 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:

• A. $12m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $8m{s}^{-1}$
• D. $10m{s}^{-1}$

Answer 1

The correct option is D $10m{s}^{-1}$

Step 1: Draw a labelled diagram.
Step 2: Find the minimum value of $u$.
Acceleration of the bus $a=1m{s}^{-2}$
Velocity of the student $u$
Initial velocity of bus $u_1=0$
Let the distance travelled by bus be $x$ when the student caught it after time $t$.
By $2^{nd}$ equation of motion,

$S=ut+\frac{1}{2}g{t}^2$

$\therefore x=\frac{1}{2}a{t}^2$

$\Rightarrow x=\frac{t^2}{2}....\left(i\right)$

$(\because a=1m{s}^{-2})$

Distance travelled by student, put $2^{nd}$ equation of motion.
$s=ut+\frac{1}{2}g{t}^2=x+50$

$\therefore x+50=ut...\left(ii\right))$

From (𝑖) and (𝑖𝑖),

$\frac{t^2}{2}+50=ut$

$\Rightarrow t^2-2ut+100=0$

The above relation must have real roots i.e., its discriminant $\ge 0$

$(2u{)}^2-4\times 100\ge 0$

$u^2-100\ge 0$

$\Rightarrow u>10m{s}^{-1}$

Thus, the minimum velocity of the student must be $10m{s}^{-1}$

Final Answer: (c)