Question

A particle moves from the point $(2.0\hat{i}+40\hat{j})m$, at $t=0$, with an initial velocity $(5.0\hat{i}+4.0\hat{j})m{s}^{-1}$. 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. $20\sqrt{2}$m
• B. $10\sqrt{2}$m
• C. $5$m
• D. $15$m

Answer 1

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

$\overrightarrow{s}=(5\hat{i}+4\hat{j})2+\frac{1}{2}(4\hat{i}+4\hat{j})4$
$=10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}$
$\overrightarrow{r_f}=\overrightarrow{r_i}=18\hat{i}+16\hat{j}$
$\overrightarrow{r_f}=20\hat{i}+20\hat{j}$
$|\overrightarrow{r_f}|=20\sqrt{2}$.