Q 11-03-039NEETJEE MainTop questionMedium
A particle moves on a circle of radius $2$ m. At a certain instant its speed is $4$ m/s and its speed is increasing at $6\ \text{m/s}^2$. The magnitude of its acceleration at that instant is
Answer: (D) $10\ \text{m/s}^2$
Centripetal (radial) acceleration: $a_c = \dfrac{v^2}{r} = \dfrac{16}{2} = 8\ \text{m/s}^2$.
Tangential acceleration: $a_t = 6\ \text{m/s}^2$.
These are perpendicular: $a = \sqrt{8^2 + 6^2} = 10\ \text{m/s}^2$.
Solution by Sreeraj P, M.Sc Physics