Kinematics Question 425
A particle moves along a path ABCD. Then the magnitude of net displacement of the particle from position A to D is:
Options:
A) 10 m
B) $ 5\sqrt{2}m $
C) 9 m
D) $ 7\sqrt{2}m $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The displacement is $ \sqrt{{{(AF)}^{2}}+{{(FD)}^{2}}}=7\sqrt{2}m $ .
BETA
