Let's solve the problem step by step:

Problem Details:

The displacement $x$ of a particle $P$ is given by: $x = kt^2 - t^3$ where $k$ is a constant, and $t$ is time in seconds.

• At $t = 2$, the velocity of the particle $P$ is $24 \, \text{m/s}$.
• The particle comes to instantaneous rest (velocity $= 0$) at point $A$.
• We are tasked to find the distance between $O$ (origin) and $A$.

Step 1: Derive velocity $v(t)$

The velocity $v(t)$ is the derivative of $x(t)$ with respect to time: $v(t) = \frac{dx}{dt} = 2kt - 3t^2$

Step 2: Use $t = 2$ to find $k$

At $t = 2$, the velocity $v(2) = 24 \, \text{m/s}$. Substituting $t = 2$ into the velocity equation: $v(2) = 2k(2) - 3(2)^2$ $24 = 4k - 12$ $4k = 36 \quad \Rightarrow \quad k = 9$

So the equation of $x(t)$ becomes: $x = 9t^2 - t^3$

The velocity equation becomes: [ v(t) =