JEE MAIN - Physics (2009 - No. 28)
A particle has an initial velocity $$3\widehat i + 4\widehat j$$ and an acceleration of $$0.4\widehat i + 0.3\widehat j$$. Its speed after 10 s is:
$$7\sqrt 2 $$ units
7 units
8.5 units
10 units
Explanation
Given $$\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s$$
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t $$
$$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$$
$$ = 7\widehat i + 7\widehat j$$
We know speed is equal to magnitude of velocity.
$$\therefore$$ $$\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$$ units
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t $$
$$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$$
$$ = 7\widehat i + 7\widehat j$$
We know speed is equal to magnitude of velocity.
$$\therefore$$ $$\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$$ units
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