A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. When does the particle stop?

The correct option is 1 second.

To find when the particle stops, we need to determine the time $t$ at which its velocity becomes zero ($v=0$).

The position of the particle is given by the displacement equation:
$x=t^3+6t^2-15t+18$

The velocity $v$ of the particle is the rate of change of displacement with respect to time, which is found by differentiating $x$ with respect to $t$:
$v=\frac{dx}{dt}$

Differentiating each term of the position function:
$v=\frac{d}{dt}\left(t^3+6t^2-15t+18\right)$
$v=3t^2+12t-15$

The particle stops when its velocity is zero:
$3t^2+12t-15=0$

We can simplify this quadratic equation by dividing the entire equation by 3:
$t^2+4t-5=0$

Factorizing the quadratic equation:
$t^2+5t-t-5=0$
$t(t+5)-1(t+5)=0$
$(t-1)(t+5)=0$

This gives two possible values for time $t$:
$t=1\text{second}$  or  $t=-5\text{seconds}$

Since time cannot be negative, we discard $t=-5$ seconds.

Thus, the particle stops at $t=1$ second.