Physics Two Dimensional Motion Question 42
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 (As clear from the figure).
BETA
