The position of a particle as a function of time $t$ is given by $x(t) = at + bt^2 - ct^3$,where $a, b,$ and $c$ are constants. When the particle attains zero acceleration,its velocity will be:

  • A
    $a + \frac{b^2}{4c}$
  • B
    $a + \frac{b^2}{c}$
  • C
    $a + \frac{b^2}{2c}$
  • D
    $a + \frac{b^2}{3c}$

Explore More

Similar Questions

The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

An object travelling at a speed of $36 \ km/h$ comes to rest in a distance of $200 \ m$ after the brakes are applied. The retardation produced by the brakes is (in $m/s^2$):

The displacement $x$ of a particle along a straight line at time $t$ is given by $x = a_0 + a_1t + a_2t^2$. The acceleration of the particle is

$A$ vehicle starts from rest and accelerates along a straight path at $2 \,m/s^2$. At the starting point of the vehicle, there is a stationary electric siren. How far has the vehicle nearly gone when the driver hears the siren at $94 \%$ of its original frequency when the vehicle was at rest (in $\,m$)? (Speed of sound $= 330 \,m/s$)

$A$ body of mass $1\,kg$ crosses a point $O$ with a velocity $60\,ms^{-1}$. $A$ force of $10\,N$ directed towards $O$ begins to act on it. It will again cross $O$ in ......... $\sec$.

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