A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)\,m$, at $t = 0$ with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1  }}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2\,s$

  • [JEE MAIN 2019]
  • A

    $15\,m$

  • B

    $20\sqrt 2 \,m$

  • C

    $5\,m$

  • D

    $10\sqrt 2 \,m$

Similar Questions

The figure shows the velocity $(v)$ of a particle plotted against time $(t)$

The initial position of an object at rest is given by $3 \hat{i}-8 \hat{j}$ It moves with constant acceleration and reaches to the position $2 \hat{i}+4 \hat{j}$ after $4 \,s$. What is its acceleration?

It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him $3$ minute to walk up . ....... $\sec$ long will it take for the passenger to arrive at the top if he walks up the moving escalator ?

Let $\vec v$ and $\vec a$ denote the velocity and acceleration respectively of a body in one-dimensional motion 

The velocity of a body at time $ t = 0$ is $10\sqrt 2 \,m/s$ in the north-east direction and it is moving with an acceleration of $ 2 \,m/s^{2}$ directed towards the south. The magnitude and direction of the velocity of the body after $5\, sec$ will be