1. Identify the Goal: The problem asks for the power delivered to the particle. Given the probable answer provided, this refers to the Average Power delivered over the time interval $T$ to achieve speed $v$.
2. Apply Work-Energy Theorem: The work done ($W$) on the particle is equal to the change in its kinetic energy ($\Delta K$).

• Initial Kinetic Energy ($K_i$) = 0 (starts from rest)
• Final Kinetic Energy ($K_f$) = $\frac{1}{2}Mv^2$ $W = \Delta K = \frac{1}{2}Mv^2 - 0 = \frac{1}{2}Mv^2$ [Class 11 Physics, Ch 5, Sec 5.2, Eq 5.2a]

3. Calculate Average Power: Average power ($P_{av}$) is defined as the work done divided by the time taken. $P_{av} = \frac{W}{t}$ Substituting the values: $P_{av} = \frac{\frac{1}{2}Mv^2}{T} = \frac{1}{2}\frac{Mv^2}{T}$ [Class 11 Physics, Ch 5, Sec 5.10, Eq 5.20]

(Note: If the question asked for Instantaneous Power at time $T$, the answer would be $P = F \cdot v = (Ma)v = M(\frac{v}{T})v = \frac{Mv^2}{T}$).