Question

A force $\overrightarrow{F}=(2\hat{i}+\hat{j}+3\hat{k})N$ acts on a particle of mass $1kg$ for $2s$. If initial velocity of particle is $u=(2\hat{j}+\hat{i}m/s$. Speed of particle at the end of $2s$ will be

• A. $12m/s$
• B. $6m/s$
• C. $9m/s$
• D. $4m/s$

Answer 1

The correct option is C $9m/s$

Given,

Initial velocity, $u=(2\hat{j}+\hat{k})m/s$

Force, $F=(2\hat{i}+\hat{j}+3\hat{k})N$

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{1}=(2\hat{i}+\hat{j}+3\hat{k})m/s^2$

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$

$\overrightarrow{v}=(2\hat{j}+\hat{k})+(2\hat{i}+\hat{j}+3\hat{k})\times 2$

$\overrightarrow{v}=(4\hat{i}+4\hat{j}+7\hat{k})m/s$

$\left|\overrightarrow{v}\right|=\sqrt{4^2+4^2+7^2}=9m/s$

Final velocity is $9m/s$