A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?

The correct option is "22cm/sec".

Step-by-Step Explanation:
We are given the position function of a particle moving in a straight line, representing the distance $x$ (in centimeters) as a function of time $t$ (in seconds):

$$x=t^3+6t^2-15t+18$$

Velocity ($v$) is defined as the rate of change of distance with respect to time. Mathematically, it is the first derivative of the position function $x$ with respect to $t$:

$$v=\frac{dx}{dt}$$

Differentiating each term of the position equation with respect to $t$ using the power rule (which states that $\frac{d}{dt}\left(t^n\right)=n{t}^{n-1}$), we get:

$$v=\frac{d}{dt}\left(t^3+6t^2-15t+18\right)$$

$$v=3t^2+12t-15$$

Now, we calculate the velocity of the particle at the end of $t=2$ seconds by substituting $t=2$ into the velocity equation:

$$v=3{\left(2\right)}^2+12\left(2\right)-15$$

$$v=3\left(4\right)+24-15$$

$$v=12+24-15$$

$$v=36-15$$

$$v=22\text{ cm/sec}$$

Therefore, the velocity of the particle at the end of 2 seconds is 22 cm/sec, which corresponds to the correct option 22cm/sec.