A particle $P$ is at the origin starts with velocity $\overrightarrow{\mathbf{v}} = \left( 2\widehat{\mathbf{i}} - 4\widehat{\mathbf{j}} \right)ms^{- 1}$ with constant acceleration $\left( 3\widehat{\mathbf{i}} - 5\widehat{\mathbf{j}} \right)ms^{- 2}$. After travelling for 2 s, its distance from the origin is

Question

Dr. Abubacker · Accepted Answer

(b) Since, acceleration is constant
$	herefore\ \overrightarrow{\mathbf{s}} = \overrightarrow{\mathbf{u}} + \frac{1}{2}\overrightarrow{\mathbf{a}}t^{2}$
$= \left( 2\widehat{i} - 4\widehat{\mathbf{j}} ight)t + \frac{1}{2}\left( 3\widehat{\mathbf{i}} + 5\widehat{\mathbf{j}} ight)t^{2}$
$= \left( 2\widehat{\mathbf{i}} - 4\widehat{j} ight)2 + \frac{1}{2}\left( 3\widehat{\mathbf{i}} + 5\widehat{\mathbf{j}} ight)2^{2}$
$= 10\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}}$
$\left| \overrightarrow{\mathbf{s}} ight| = \sqrt{10^{2} + 2^{2}} = \sqrt{104} = 10.2m$

Explanation

Reviewed by Dr. Abubacker

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