Kinematics Question 366

Question: 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) zero

Show Answer

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}+0} $

$ [ As|v _{1}|=|v _{2}|=5m/s ]=5\sqrt{2}m/s $

$ Avg.acc.=\frac{\Delta }{t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/s^{2} $
$ \Rightarrow \tan \theta =\frac{5}{-5}=-1 $ Which means $ \theta $ it’s in the second quadrant. (Towards north-west)



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक