Kinematics Question 336
Question: The displacement x of a particle at the instant when it’s velocity is v is given by $ v=\sqrt{3x+16}. $ it’s acceleration and initial velocity are
Options:
A) 1.5 unit, 4 unit
B) 3 unit, 4 unit
C) 16 unit, 1.6 unit
D) 16 unit, 3 unit
Show Answer
Answer:
Correct Answer: A
Solution:
$ \text{v=}\sqrt{3x+16}\Rightarrow {{v}^{2}}=3x+16 $
$ \Rightarrow v^{2}-16=3x $
Comparing with $ v^{2}-u^{2}=2aS $ , we get, $ u=4 $ unit, $ 2a=3 $ or $ a=1.5unit $
BETA
