Kinematics Question 366
A particle is moving eastwards with a velocity of $ 5m{{s}^{-1}} $ . In 10 seconds the velocity changes to $ 5m{{s}^{-1}} $ northwards. The average acceleration in this time is
Options:
A) $ \frac{1}{2}m{{s}^{-2}} $ toward north
B) $ \frac{1}{\sqrt{2}}m{{s}^{-2}} $ toward north-east
C) $ \frac{1}{\sqrt{2}}m{{s}^{-2}} $ towards north-west
D) 0
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Answer:
Correct Answer: C
Solution:
$ \text{Average acceleration =}\frac{\text{change in velocity}}{\text{time interval}} $
$ \text{=}\frac{\Delta \overrightarrow{v}}{t} $
$ \overrightarrow{v _{1}}=5\hat{i},\overrightarrow{v _{2}}=5\hat{j} $
$ \Delta \overrightarrow{v}=( \overrightarrow{v _{2}}-\overrightarrow{v _{1}} ) $
$ =\sqrt{v _{1}^{2}+v _{2}^{2}+2v _{1}v _{2}\cos 90} $
$ =\sqrt{5^{2}+5^{2}}$
$ [ As|v _{1}|=|v _{2}|=5m/s ]=5\sqrt{2}m/s $
$ Avg.acc.=\frac{\Delta v}{t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/s^{2} $ $ \Rightarrow \tan \theta =\frac{5}{-5}=-1 $ Which means $ \theta $ is in the second quadrant. (Towards north-west)
BETA
