Question

A particle moves from the point $\left( 2.0\widehat{i}+4.0\widehat{j} \right)m$ at t = 0 with an initial velocity $\left( 5.0\widehat{i}+4.0\widehat{j} \right)m{{s}^{-1}}$ . It is acted upon by a constant force which produces a constant acceleration$\left( 4.0\widehat{i}+4.0\widehat{j} \right)m{{s}^{-2}}$ . What is the distance of the particle from the origin at time 2s?
A. $20\sqrt{2}m$
B. $15m$
C. $10\sqrt{2}m$
D. $5m$

Answer 1

From the given initial velocity vector and acceleration vector you could find the displacement vector of the given particle at the end of 2s. We know that the difference of final position vector and initial position vector will give us the displacement vector. Thus we will get the final position and its magnitude will give us the required distance.
Formula used:
Newton’s equation of motion,
$\overrightarrow{s}=\overrightarrow{u}t+\dfrac{1}{2}\overrightarrow{a}{{t}^{2}}$

Complete answer:
In the question we are given the initial position vector of a particle at t = 0 as,
$\overrightarrow{{{r}_{i}}}=\left( 2.0\widehat{i}+4.0\widehat{j} \right)m$
The initial velocity of the particle is given as,
$\overrightarrow{u}=\left( 5.0\widehat{i}+4.0\widehat{j} \right)m{{s}^{-1}}$
A constant force is also being exerted on the given particle. The constant acceleration produced by the force is given as,
$\overrightarrow{a}=\left( 4.0\widehat{i}+4.0\widehat{j} \right)m{{s}^{-2}}$
We are supposed to find the distance of the particle from the origin at time 2s. Let $\overrightarrow{{{r}_{f}}}$ be the position vector of the particle at time t = 2s. Firstly, let us find the displacement vector from the initial position at t =0s to the final position at t= 2s.
Newton’s equations of motion in the vector form is given by,
$\overrightarrow{s}=\overrightarrow{u}t+\dfrac{1}{2}\overrightarrow{a}{{t}^{2}}$
$\Rightarrow \overrightarrow{s}=\left( 5.0\widehat{i}+4.0\widehat{j} \right)\left( 2 \right)+\dfrac{1}{2}\left( 4.0\widehat{i}+4.0\widehat{j} \right){{\left( 2 \right)}^{2}}$
$\Rightarrow \overrightarrow{s}=10\widehat{i}+8.0\widehat{j}+8\widehat{i}+8\widehat{j}$
$\therefore \overrightarrow{s}=\left( 18\widehat{i}+16\widehat{j} \right)m$
We know that the displacement vector can be given by,
$\overrightarrow{{{r}_{f}}}-\overrightarrow{{{r}_{i}}}=\overrightarrow{s}$
So the final position vector could be given by,
$\overrightarrow{{{r}_{f}}}=\overrightarrow{s}+\overrightarrow{{{r}_{i}}}$
$\Rightarrow \overrightarrow{{{r}_{f}}}=\left( 18\widehat{i}+16\widehat{j} \right)+\left( 2.0\widehat{i}+4.0\widehat{j} \right)$
$\therefore \overrightarrow{{{r}_{f}}}=20\widehat{i}+20\widehat{j}$
Now, the distance of the particle from the origin could be given by the magnitude of the final position vector at t=2s,
$\left| \overrightarrow{{{r}_{f}}} \right|=\sqrt{{{20}^{2}}+{{20}^{2}}}$
$\therefore \left| \overrightarrow{{{r}_{f}}} \right|=20\sqrt{2}m$
Therefore, we found the distance of the particle from the origin at time 2s as$20\sqrt{2}m$.

Hence, option A is the right answer.

Note:
The position vector of a particle determines the position of the body relative to a point (origin) in the coordinate system. When the point moves, the position vector also changes in both magnitude as well as direction. The time rate of change of position vector will give us the velocity vector.