$A$ particle moves from the point $(2.0\hat i + 4.0\hat j) \, m$ at $t = 0$ with an initial velocity $(5.0\hat i + 4.0\hat j) \, m/s$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat i + 4.0\hat j) \, m/s^2$. What is the distance of the particle from the origin at time $t = 2 \, s$?

  • A
    $15 \, m$
  • B
    $20\sqrt{2} \, m$
  • C
    $5 \, m$
  • D
    $10\sqrt{2} \, m$

Explore More

Similar Questions

$A$ body $P$ is projected at an angle of $30^{\circ}$ with the horizontal and another body $Q$ is projected at an angle of $30^{\circ}$ with the vertical. If the ratio of the horizontal ranges of the bodies $P$ and $Q$ is $1: 2$, then the ratio of the maximum heights reached by the bodies $P$ and $Q$ is

$A$ particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y \hat{i} + 2x \hat{j})$,where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

$A$ scooter going due east at $10\, ms^{-1}$ turns right through an angle of $90^\circ$. If the speed of the scooter remains unchanged in taking the turn,the change in the velocity of the scooter is

Difficult
View Solution

At any time $t$,the coordinates of a moving particle are $x = at^2$ and $y = bt^2$. The speed of the particle is

$A$ particle moves along a circle of radius $R$ with kinetic energy $K = as^2$,where $s$ is the distance traveled. What is the net force acting on the particle?

Difficult
View Solution

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