JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 3)
A force $$\overrightarrow F = (40\widehat i + 10\widehat j)N$$ acts on a body of mass 5 kg. If the body starts from rest, its position vector $$\overrightarrow r $$ at time t = 10 s, will be :
$$(100\widehat i + 400\widehat j)m$$
$$(100\widehat i + 100\widehat j)m$$
$$(400\widehat i + 100\widehat j)m$$
$$(400\widehat i + 400\widehat j)m$$
Explanation
$${{d\overrightarrow v } \over {dt}} = \overrightarrow a = {{\overrightarrow F } \over m} = (8\widehat i + 2\widehat j)m/{s^2}$$
$${{d\overrightarrow r } \over {dt}} = \overrightarrow v = (8t\widehat i + 2t\widehat j)m/s$$
$$\overrightarrow r = (8\widehat i + 2\widehat j){{{t^2}} \over 2}m$$
At t = 10 sec
$$\overrightarrow r = \left[ {(8\widehat i + 2\widehat j)50} \right]m$$
$$ \Rightarrow \overrightarrow r = (400\widehat i + 100\widehat j)m$$
$${{d\overrightarrow r } \over {dt}} = \overrightarrow v = (8t\widehat i + 2t\widehat j)m/s$$
$$\overrightarrow r = (8\widehat i + 2\widehat j){{{t^2}} \over 2}m$$
At t = 10 sec
$$\overrightarrow r = \left[ {(8\widehat i + 2\widehat j)50} \right]m$$
$$ \Rightarrow \overrightarrow r = (400\widehat i + 100\widehat j)m$$
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