The motion of a particle is described by the equation $x=a+b{t}^2$, where $a=15cm$ and $b=3cm/s^2$. Its instantaneous velocity at a time $3$ sec will be.

The correct option is B

$18cm/sec$

Step 1: Given data

The motion of particles is given as

$x=a+b{t}^2$

Value of $a=15cm$

Value of $b=3cm/s^2$

Time $t=3sec$

Step 2: Formula used

Velocity is defined as the rate of change of displacement.

$velocity\left(v\right)=\frac{dx}{dt}$

Where $x$ is the displacement

Step 3: Compute the velocity

The given equation of motion of a particle is $x=a+b{t}^2····\left(1\right)$

So, differentiate the equation $\left(1\right)$ with respect to time to obtain the velocity

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d\left(a+b{t}^2\right)}{dt}\\ \frac{dx}{dt}& =& 0+2bt\\ v& =& 2bt·····\left(2\right)\end{array}$

Substitute the value of $b=3$ in equation $\left(2\right)$ to obtain the instantaneous velocity at a time $t=3$ sec

$$\begin{gathered} v=2\times 3\times t \\ v=6t\dots \dots \dots \dots \dots \left(3\right) \end{gathered}$$

Now substitute the value of $t=3sec$ in equation (3)

$\begin{array}{rcl}v& =& 6\times 3\\ & =& 18\frac{cm}{s}\end{array}$

Therefore, at the time $t=3sec$, its instantaneous velocity will be $18cm/sec$.

Hence, option B is the correct answer.