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 acceleration of the particle at the end of 2 seconds?

The correct option is 24cm/sec².

To find the acceleration of the particle, we need to understand the relationship between distance, velocity, and acceleration with respect to time.

First, let's write down the given equation for the distance, $x$ (in cm), as a function of time, $t$ (in seconds):

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

Velocity ($v$) is the first derivative of distance ($x$) with respect to time ($t$):

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

Differentiating the expression for $x$ with respect to $t$, we get:

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

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

Acceleration ($a$) is the first derivative of velocity ($v$) with respect to time ($t$), or the second derivative of distance ($x$) with respect to time ($t$):

$$a=\frac{dv}{dt}$$

Differentiating the expression for $v$ with respect to $t$, we get:

$$a=\frac{d}{dt}\left(3t^2+12t-15\right)$$

$$a=6t+12$$

Now, we need to calculate the acceleration at the end of 2 seconds, which means we substitute $t=2$ into the acceleration equation:

$$a=6\left(2\right)+12$$

$$a=12+12$$

$$a=24{\text{cm/sec}}^2$$

Thus, the acceleration of the particle at the end of 2 seconds is indeed 24cm/sec².