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

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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 $



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