Kinematics Question 495

Question: 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[IIT 1982; AFMC 1999; Pb PET 2000; JIPMER 2001, 02]

Options:

A) Zero

B) $ \frac{1}{\sqrt{2}}m\text{/}{{s}^{\text{2}}} $ toward north-west

C) $ \frac{1}{\sqrt{2}}m\text{/}{{s}^{\text{2}}} $ toward north-east

D) $ \frac{1}{2}m\text{/}{{s}^{\text{2}}} $ toward north-west

Show Answer

Answer:

Correct Answer: B

Solution:

$ \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}}\text{m/}{{\text{s}}^{\text{2}}} $

Directed toward north-west



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