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
BETA
