Let the tangential acceleration of the particle be $a_t$. Given: Initial velocity, $u = 0$ (starts from rest) Final velocity, $v = 80 \text{ m/s}$ Radius, $r = \frac{20}{\pi} \text{ m}$ Number of revolutions, $n = 2$

The total linear distance $s$ covered by the particle along the circular path in 2 revolutions is: $s = n \times 2\pi r = 2 \times 2\pi \times \left(\frac{20}{\pi}\right) = 80 \text{ m}$

Using the third equation of motion ($v^2 = u^2 + 2a_ts$): $(80)^2 = 0^2 + 2 \times a_t \times 80$ $6400 = 160 a_t$ $a_t = \frac{6400}{160} = 40 \text{ m/s}^2$.