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किसी कण का प्रारंभिक वेग $(2 \vec{i}+3 \vec{j})$ तथा त्वरण $(0.3 \vec{i}+0.2 \vec{j})$ है। $10$ सेकण्ड बाद कण के वेग का मान होगा
$9 \sqrt 2 $ मात्रक
$5\sqrt 2 $ मात्रक
$5 $ मात्रक
$9 $ मात्रक
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,Here,\\
Initial\,velocity,\,\overline u = 2\hat i + 3\hat j\\
Acceleration,\,\overline a = 0.3\hat i + 0.2\hat j\\
Time\,,\,t = 10\,s\\
Let\,\overline v \,be\,velocity\,of\,a\,particle\,\\
after\,10\,s.
\end{array}$
$\begin{array}{l}
{\rm{Using,}}\,\overline v \, = \overline u + \,\overline a t\\
\therefore \,\,\,\overline v = \left( {2\hat i + 3\hat j} \right) + \left( {0.3\hat i + 0.2\hat j} \right)\left( {10} \right)\\
\,\,\,\,\,\,\,\,\, = \,2\hat i + 3\hat j + 3\hat i + 2\hat j = 5\hat i + 5\hat j\\
Speed\,of\,the\,particle\,after\,10\,s = \left| {\overline v } \right|\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( 5 \right)}^2} + {{\left( 5 \right)}^2}} = 5\sqrt 2 \,units
\end{array}$