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A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)\,m$, at $t = 0$ with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1 }}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2\,s$
$15\,m$
$20\sqrt 2 \,m$
$5\,m$
$10\sqrt 2 \,m$
Solution
$\begin{array}{l}
\overrightarrow S = \left( {5\hat i + 4\hat j} \right)2 + \frac{1}{2}\left( {4\hat j + 4\hat j} \right)4\\
\,\,\,\, = \,10\hat i + 8\hat j + 8\hat i + 8\hat j\\
\overrightarrow {{r_2}} – \overrightarrow {{r_1}} = 18\hat i + 16\hat j\\
\overrightarrow {{r_2}} = 20\hat i + 20\hat j\\
\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2
\end{array}$
