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3-2.Motion in Plane
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A particle has an initial velocity of ($3\hat i + 4\hat j)\;ms^{-1}$ and an acceleration of $(0.4\hat i + 0.3\hat j)\;ms^{-2}$ Its speed after $10\;s$ is:
A$7 $ unit
B$8.5$ unit
C$10$ unit
D$7\sqrt 2 $ unit
(AIEEE-2009) (AIPMT-2010)
Solution
$\begin{array}{l}
Given\,\vec u = 3\hat i + 4\hat j,\hat a = 0.4\hat i + 0.3\hat j,t = 10s\\
\vec v = \vec u + \vec at = 3\hat i + 4\hat j + \left( {0.4\hat i + 0.3\hat j} \right) \times 10\\
= 7\hat i + 7\hat j\\
\therefore \left| {\vec v} \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,units
\end{array}$
Given\,\vec u = 3\hat i + 4\hat j,\hat a = 0.4\hat i + 0.3\hat j,t = 10s\\
\vec v = \vec u + \vec at = 3\hat i + 4\hat j + \left( {0.4\hat i + 0.3\hat j} \right) \times 10\\
= 7\hat i + 7\hat j\\
\therefore \left| {\vec v} \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,units
\end{array}$
Standard 11
Physics
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