2}}+{{a}^{2}}{{t}^{2}}+2u\,at\cos \theta }\]                         \[v=\sqrt{200+100+2\times 10\sqrt{2}\times 10\times \cos 135}\]\[=10\,m/s\] \[\TAN \ALPHA =\frac{at\sin \theta }{u+at\cos \theta }\frac{10\sin 135}{10\sqrt{2}+10\cos 135}=1\]\\[\alpha =45{}^\circ \]                         i.e. resultant VELOCITY is 10 m/s towards East.

"> 2}}+{{a}^{2}}{{t}^{2}}+2u\,at\cos \theta }\]                         \[v=\sqrt{200+100+2\times 10\sqrt{2}\times 10\times \cos 135}\]\[=10\,m/s\] \[\TAN \ALPHA =\frac{at\sin \theta }{u+at\cos \theta }\frac{10\sin 135}{10\sqrt{2}+10\cos 135}=1\]\\[\alpha =45{}^\circ \]                         i.e. resultant VELOCITY is 10 m/s towards East.

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The velocity of a body at time t = 0 is \[10\sqrt{2}\] m/s in the north-east direction and it is moving with an acceleration of 2 m/s2 directed towards the south.  The magnitude and direction of the velocity of the body after 5 sec will be                                     [AMU (Engg.) 1999]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 9 months ago

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\[\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}\,t\]\ \[v=\sqrt{{{u}^{2}}+{{a}^{2}}{{t}^{2}}+2u\,at\cos \theta }\]                         \[v=\sqrt{200+100+2\times 10\sqrt{2}\times 10\times \cos 135}\]\[=10\,m/s\] \[\TAN \ALPHA =\frac{at\sin \theta }{u+at\cos \theta }\frac{10\sin 135}{10\sqrt{2}+10\cos 135}=1\]\\[\alpha =45{}^\circ \]                         i.e. resultant VELOCITY is 10 m/s towards East.

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Posted on 14 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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