Question

A particle starts from the origin at $t=0$ with a velcoity of $5\hat{i}m/s$ and moves in $x-y$ plane with a constant acceleration of $(4\hat{i}+2\hat{j})m/s^2$. What is the speed of the particle at the time $t=3s$?

• A. $18m/s$
• B. $16m/s$
• C. $19m/s$
• D. $20m/s$

Answer 1

The correct option is A $18m/s$

Given, initial velocity of the particle,
$\overrightarrow{u}=5\hat{i}m/s$
And acceleration of the particle,
$\overrightarrow{a}=(4\hat{i}+2\hat{j})m/s^2$
Velocity of the particle after $t=3s$,
$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
$\overrightarrow{v}=5\hat{i}+(4\hat{i}+2\hat{j})\left(3\right)$
$\overrightarrow{v}=(17\hat{i}+6\hat{j})m/s$
So, speed of the particle,
$v=\sqrt{(17{)}^2+(6{)}^2}$
$v=\sqrt{325}=18m/s$