A particle has initial (t = 0) velocity $\stackrel{\to <!--\; \to \; -->}{\mathrm{\mu <!--\; \mu \; -->}}=5\stackrel{^<!--\; ^\; -->}{i}$ and is at origin at this instant. Its acceleration is given by $(3\stackrel{^<!--\; ^\; -->}{i}+4\stackrel{^<!--\; ^\; -->}{j}).$ When particle’s x co-ordinate is 16 units, then its speed is

The correct option is √185 units.

Let us solve the problem step-by-step.

1. Identify the given parameters:
The initial velocity vector of the particle at time $t=0$ is:
$\overrightarrow{u}=5\hat{i}$
This gives the initial components of velocity along the x and y axes as:
$u_x=5\text{units}$
$u_y=0\text{units}$

The particle is initially at the origin, which means:
$x_0=0$
$y_0=0$

The constant acceleration vector is:
$\overrightarrow{a}=3\hat{i}+4\hat{j}$
This gives the components of acceleration as:
$a_x=3{\text{units/s}}^2$
$a_y=4{\text{units/s}}^2$

2. Find the time when the x-coordinate of the particle is 16 units:
Using the second equation of motion for the x-direction:
$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Substitute the given values:
$16=5t+\frac{1}{2}\left(3\right)t^2$
$16=5t+1.5t^2$

Multiplying the entire equation by 2 to clear the fraction:
$32=10t+3t^2$
$3t^2+10t-32=0$

Solving this quadratic equation for $t$ using the quadratic formula:
$t=\frac{-10\pm \sqrt{{10}^2-4\left(3\right)(-32)}}{2\left(3\right)}$
$t=\frac{-10\pm \sqrt{100+384}}{6}$
$t=\frac{-10\pm \sqrt{484}}{6}$
$t=\frac{-10\pm 22}{6}$

Since time $t$ must be positive:
$t=\frac{-10+22}{6}=\frac{12}{6}=2\text{seconds}$

3. Determine the velocity components at $t=2\text{seconds}$:
For the x-component of velocity:
$v_x=u_x+a_{x}t$
$v_x=5+3\left(2\right)=11\text{units/s}$

For the y-component of velocity:
$v_y=u_y+a_{y}t$
$v_y=0+4\left(2\right)=8\text{units/s}$

4. Calculate the speed of the particle:
The speed $v$ is the magnitude of the velocity vector:
$v=\sqrt{v_x^2+v_y^2}$
$v=\sqrt{{11}^2+8^2}$
$v=\sqrt{121+64}$
$v=\sqrt{185}\text{units}$