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3-2.Motion in Plane
normal
A particle is moving eastwards with velocity of $5\,m/s$. In $10 \,sec$ the velocity changes to $5 \,m/s$ northwards. The average acceleration in this time is
A
Zero
B
$\frac{1}{{\sqrt 2 }}\,\,m{\rm{/}}{s^{\rm{2}}}$ toward north-west
C
$\frac{1}{{\sqrt 2 }}\,\,m{\rm{/}}{s^{\rm{2}}}$ toward north-east
D
$\frac{1}{2}\,\,m{\rm{/}}{s^{\rm{2}}}$toward north-west
Solution
(b) $\Delta \vec \upsilon = {\vec \upsilon _2} – {\vec \upsilon _1}$$ = \sqrt {\upsilon _1^2 + \upsilon _2^2 – 2{\upsilon _1}{\upsilon _2}\,\,\cos {{90}^o}} $$ = \sqrt {{5^2} + {5^2}} = 5\sqrt 2 $
Average acceleration
$ = \frac{{\Delta \upsilon }}{{\Delta t}} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}\,\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$
Directed toward north-west (As clear from the figure).
Standard 11
Physics
