Question

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

• A. $1.5$ units, $4$ units
• B. $3$ units, $4$ units
• C. $16$ units, $1.6$ units
• D. $16$ units, $3$ units

Answer 1

The correct option is A $1.5$ units, $4$ units

WKT, $3^{rd}$ equation of Kinematics is given by $v^2=u^2+2aS--\left(1\right)$,
Where, $u=$ initial velocity
$v=$ final velocity , $a=$ acceleration and $S=$ Distance
Given , $v=\sqrt{3x+16}$
Squaring both side we get, $v^2=3x+16\Rightarrow v^2-16=3x\Rightarrow v^2-4^2=3x--\left(2\right)$
Now, Comparing equation $\left(1\right)$ and $\left(2\right)$ , we get
$u=4units$ and $2a=3\Rightarrow a=1.5units$