A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled measured along the circumference is given by S= t²/2 + t³/3. The acceleration of particle when t = 2 sec is

To find the acceleration of a particle moving along a circular path, we must consider both its tangential acceleration ($a_t$) and its centripetal (or radial) acceleration ($a_c$). The total acceleration ($a$) is the vector sum of these two perpendicular components:
$a=\sqrt{{a_t}^2+{a_c}^2}$

Step 1: Find the velocity and tangential acceleration
The distance travelled along the circumference is given by:
$S=\frac{t^2}{2}+\frac{t^3}{3}$
The speed $v$ of the particle is the rate of change of distance with respect to time:
$v=\frac{dS}{dt}=\frac{d}{dt}\left(\frac{t^2}{2}+\frac{t^3}{3}\right)=t+t^2$
The tangential acceleration $a_t$ is the rate of change of speed:
$a_t=\frac{dv}{dt}=\frac{d}{dt}\left(t+t^2\right)=1+2t$

Step 2: Calculate the values at time $t=2$ seconds
At $t=2$ s, the speed $v$ is:
$v=2+2^2=2+4=6\text{m/s}$
At $t=2$ s, the tangential acceleration $a_t$ is:
$a_t=1+2\left(2\right)=1+4=5{\text{m/s}}^2$

Step 3: Calculate the centripetal acceleration
The centripetal acceleration $a_c$ is given by the formula:
$a_c=\frac{v^2}{R}$
Given the radius of the circular path $R=3$ m, we substitute the velocity at $t=2$ s:
$a_c=\frac{6^2}{3}=\frac{36}{3}=12{\text{m/s}}^2$

Step 4: Find the total acceleration
Now, combine the tangential and centripetal accelerations:
$a=\sqrt{{a_t}^2+{a_c}^2}=\sqrt{5^2+{12}^2}$
$a=\sqrt{25+144}=\sqrt{169}=13{\text{m/s}}^2$

Therefore, the total acceleration of the particle at $t=2$ seconds is 13 m/s².