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 :
Solution
$${{d\overrightarrow v } \over {dt}} = \overrightarrow a = {{\overrightarrow F } \over m} = (8\widehat i + 2\widehat j)m/{s^2}$$<br><br>$${{d\overrightarrow r } \over {dt}} = \overrightarrow v = (8t\widehat i + 2t\widehat j)m/s$$<br><br>$\overrightarrow r = (8\widehat i + 2\widehat j){{{t^2}} \over 2}m$<br><br>At t = 10 sec<br><br>$\overrightarrow r = \left[ {(8\widehat i + 2\widehat j)50} \right]m$<br><br>$\Rightarrow \overrightarrow r = (400\widehat i + 100\widehat j)m$
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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