Question

A particle has initial velocity $(3\hat{i}+4\hat{j})m/s$ and has acceleration $(0.4\hat{i}+0.3\hat{j})m/s^2$ $.$ Calculate its speed after $10s.$

Answer 1

We have,

$\overrightarrow{u}=(3\hat{i}+4\hat{j})m{s}^{-1}$

and $\overrightarrow{a}=(0.4\hat{i}+0.3\hat{j})m{s}^{-2}$

From I equation of motion after $10s$

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

$=(3\hat{i}+4\hat{j})+(0.4\hat{i}+0.3\hat{j})\times 10$

$=(3\hat{i}+4\hat{j})+(4\hat{i}+3\hat{j})$

$\Rightarrow \overrightarrow{v}=7\hat{i}+7\hat{j}$

$\Rightarrow |\overrightarrow{v}|=\sqrt{7^2+7^2}$

$\Rightarrow |\overrightarrow{v}|=7\sqrt{2}m{s}^{-1}$