The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time in seconds. If the particle starts with an initial velocity $v = 2\,m/s$ at $t = 0$,then the velocity at the end of $t = 2\,s$ is $............m/s$.

  • A
    $12$
  • B
    $14$
  • C
    $16$
  • D
    $18$

Explore More

Similar Questions

$A$ car accelerates from rest at a constant rate of $2 \ m/s^2$ for some time. Then it retards at a constant rate of $4 \ m/s^2$ and comes to rest. If it remains in motion for $3 \ s$,then the total displacement covered by it is $.... \ m$.

$A$ particle starts moving along a line from zero initial velocity and comes to rest after moving a distance $d$. During its motion,it has a constant acceleration $f$ over $2/3$ of the distance and covers the rest of the distance with constant retardation. The total time taken to cover the distance is

If a body having initial velocity zero is moving with uniform acceleration $8\,m/s^2$,the distance travelled by it in the fifth second will be.........$m$.

$A$ small block slides down on a smooth inclined plane,starting from rest at time $t=0$. Let $S_{n}$ be the distance travelled by the block in the interval $t=n-1$ to $t=n$. Then,the ratio $\frac{S_{n}}{S_{n+1}}$ is

The distance travelled by a particle is proportional to the square of time. Then,the particle travels with:

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