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3-2.Motion in Plane
medium
किसी कण का स्थिति सदिश $\mathop r\limits^ \to = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$ द्वारा प्रदर्शित है प्रथम $10$ सैकण्ड में तय की गई दूरी ......... $m$ है
A$500 $
B$300$
C$150$
D$100$
Solution
(a) $\mathop r\limits^ \to = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$
$t = 0$ पर, ${\overrightarrow {\;r} _1} = 7\hat k$
$t = 10\sec $ पर, ${\vec r_2} = 300\hat i + 400\hat j + 7\hat k$,
$\mathop {\Delta r}\limits^ \to = {\mathop r\limits^ \to _2} – {\mathop r\limits^ \to _1} = 300\hat i + 400\hat j$
$|\mathop {\Delta r}\limits^ \to |\, = \,|{\mathop r\limits^ \to _2} – {\mathop r\limits^ \to _1}|\, = \sqrt {{{(300)}^2} + {{(400)}^2}} = 500\,m$
$t = 0$ पर, ${\overrightarrow {\;r} _1} = 7\hat k$
$t = 10\sec $ पर, ${\vec r_2} = 300\hat i + 400\hat j + 7\hat k$,
$\mathop {\Delta r}\limits^ \to = {\mathop r\limits^ \to _2} – {\mathop r\limits^ \to _1} = 300\hat i + 400\hat j$
$|\mathop {\Delta r}\limits^ \to |\, = \,|{\mathop r\limits^ \to _2} – {\mathop r\limits^ \to _1}|\, = \sqrt {{{(300)}^2} + {{(400)}^2}} = 500\,m$
Standard 11
Physics
