$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

  • A
    $\frac{-2}{x^3}$
  • B
    $-\frac{1}{4x^3}$
  • C
    $-\frac{1}{4x^2}$
  • D
    $\frac{1}{x^2}$

Explore More

Similar Questions

$A$ car,starting from rest,accelerates at the rate $\alpha$ through a distance $d$,then continues at a constant speed for time $t$ and then decelerates at the rate of $\alpha/2$ to come to rest. If the total distance traveled is $15\, d$,then $d=$

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

Two points move in the same straight line starting at the same moment from the same point. The first moves with constant velocity $u$ and the second with constant acceleration $f$. During the time that elapses before the second catches the first,the greatest distance between the particles is $........$

Difficult
View Solution

The displacement of a particle moving in a straight line is given by the expression $x = A t^3 + B t^2 + C t + D$ in metres, where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is

The displacement $(x)$ of a particle depends on time $t$ as $x = \alpha t^2 - \beta t^3$. Choose the incorrect statements from the following.

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