Question

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 $S = \frac{t^{2}}{2} + \frac{t^{3}}{3}$. The acceleration of particle when $t = 2\mspace{6mu}\sec$is

• A. 1.3 m/s²
• B. 13 m/s²
• C. 3 m/s²
• D. 10 m/s²

Answer 1

Answer: (B)

13 m/s²

Answer 2

$s = \frac{t^{2}}{2} + \frac{t^{3}}{3}$ $\Rightarrow$ $v = \frac{ds}{dt} = t + t^{2}$

and $a_{t} = \frac{dv}{dt} = \frac{d}{dt}(t + t^{2})$

$= 1 + 2t$

At t = 2 sec, v = 6 m/s and $a_{t} = 5m/s^{2}$,

$a_{c} = \frac{v^{2}}{r} = \frac{36}{3} = 12m ⥂ / ⥂ s^{2}$

$a_{N} = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(12)^{2} + (5)^{2}} = 13m/s^{2}$.