The displacement $x$ of a particle at the instant when its velocity $v$ is given by $v = \sqrt{3x + 16}.$ Its acceleration and initial velocity are

Question

Dr. Abubacker · Accepted Answer

(a) $v = \sqrt{3x + 16}$ or $v^{2} = 3x + 16$ or $v^{2} - 16 = 3x$
Comparing with $v^{2} - u^{2} = 2aS,$ we get, $u = 4$ units , $2a = 3$
or $a = 1.5$ units.

Q. The displacement $x$ of a particle at the instant when its velocity $v$ is given by $v = \sqrt{3x + 16}.$ Its acceleration and initial velocity are

Explanation

Reviewed by Dr. Abubacker

Want more practice?