$A$ student is at a distance of $16 \ m$ from a bus when the bus begins to move with a constant acceleration of $9 \ m \ s^{-2}$. The minimum velocity with which the student should run towards the bus so as to catch it is $\alpha \sqrt{2} \ m \ s^{-1}$. The value of $\alpha$ is

  • A
    $10$
  • B
    $12$
  • C
    $15$
  • D
    $20$

Explore More

Similar Questions

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

$A$ particle is moving with speed $v = b\sqrt{x}$ along the positive $x$-axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at the origin at $t = 0$).

$A$ body starts from rest and acquires a velocity of $10 \ m \ s^{-1}$ in $2 \ s$. What is the acceleration of the body and the distance travelled?

$A$ train moves from rest with a uniform acceleration $a$. After attaining a maximum speed $v$,it starts moving with uniform retardation $a$. Assuming $s$ is the total distance covered in the unidirectional motion of the train,find its total time of journey and maximum speed.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo