A particle moves from the point $(2 .0 \hat{i} + 4 .0 \hat{j}) m$ at t = 0 with an initial velocity $(5 .0 \hat{i} + 4 .0 \hat{j}) {ms}^{- 1}$ . It is acted upon by a constant force which produces a constant acceleration $(4 .0 \hat{i} + 4 .0 \hat{j}) {ms}^{- 2}$ . What is the distance (in m) of the particle from the origin at time 2s?

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answer is 28.28.

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Detailed Solution

$\vec{S} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}; \Rightarrow \vec{S} = (5 \hat{i} + 4 \hat{j}) 2 + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) 4 = 10 \hat{i} + 8 \hat{j} + 8 \hat{i} + 8 \hat{j}$
${\vec{r}}_{f} - {\vec{r}}_{i} == 18 \hat{i} + 16 \hat{j}$

[as $\vec{s}$ = change in position = ${\vec{r}}_{f} - {\vec{r}}_{i}$ ]

${\vec{r}}_{r} = 20 \hat{i} + 20 \hat{j}$
$|{\vec{r}}_{r}| = 20 \sqrt{2} = 28.284$

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