Question

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 $2s$ ?

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

Answer 1

The correct option is D $20\sqrt{2}m$

1. Find displacement of the partical at time $2s$

$\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}$ $\overrightarrow{a}{t}^2$

Given,

initial velocity, $\overrightarrow{u}=(5.0\hat{i}+4.0\hat{j})m/s$

Acceleration, $\overrightarrow{a}=(4.0\hat{i}+4.0\hat{j}m/s^2$

So, position of the particle after $2s$,

Using, $\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$

$\overrightarrow{s}=(5.0\hat{i}+4.0\hat{j})\times 2+\frac{1}{2}(4.0\hat{i}+4.0\hat{j})\times (2{)}^2$

$\overrightarrow{s}=18\hat{i}+16\hat{j}$

2. Find distance of the particle from the origin at time $2s$.

Given, initial position of the particle,

${\overrightarrow{r}}_i=(2.0\hat{i}+4.0\hat{j})m$

Let the position of the particle at time $2s$ be ${\overrightarrow{r}}_f$.

${\overrightarrow{r}}_f-{\overrightarrow{r}}_i=\overrightarrow{s}=18\hat{i}+16\hat{j}$

${\overrightarrow{r}}_f=(18\hat{i}+16\hat{j})+(2.0\hat{i}+4.0\hat{j})$

${\overrightarrow{r}}_f=20\hat{i}+20\hat{j}$

So, distance of the particle from the origin,

$\left|{\overrightarrow{r}}_f\right|=\sqrt{(20{)}^2+(20{)}^2}$

$\left|{\overrightarrow{r}}_f\right|=20\sqrt{2}m$

Final answer : (a)