JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 8)
A particle moves from the point $$\left( {2.0\widehat i + 4.0\widehat j} \right)$$ m, at t = 0, with an initial velocity $$\left( {5.0\widehat i + 4.0\widehat j} \right)$$ ms$$-$$1. It is acted upon by a constant force which produces a constant acceleration $$\left( {4.0\widehat i + 4.0\widehat j} \right)$$ ms$$-$$2. What is the distance of the particle from the origin at time 2 s?
15 m
$$20\sqrt 2 $$ m
$$10\sqrt 2 $$ m
5 m
Explanation
$$\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4$$
$$ = 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$$
$$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$$
$$\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j$$
$$\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2 $$
$$ = 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$$
$$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$$
$$\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j$$
$$\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2 $$
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