Question

A particle is travelling in a circular path with constant speed 10 m/s. It traverses angle 53° between the positions A and B in time 2 seconds. The magnitude of average acceleration vector between the position A and B is

• A. 0
• B. 10m/s²
• C. 4√2 m/s²
• D. 2√5 m/s²

Answer 1

Answer: (A)

0

Answer 2

Total acceleration will make angle π/4 with radius when tangential acceleration = centripetal acceleration.

Total acceleration will make angle $\pi / 4$ with radius when tangential acceleration $=$ centripetal acceleration. $R \alpha = \frac { V ^ { 2 } } { R } \Rightarrow R \alpha = \frac { \left( a _ { t } t \right) ^ { 2 } } { R }$ $\Rightarrow \alpha = \frac { R ^ { 2 } \alpha ^ { 2 } \cdot t ^ { 2 } } { R ^ { 2 } }$ $\Rightarrow \alpha = \frac { 1 } { t ^ { 2 } }$

Rα = V²/R ⇒ Rα = (a_tt)²/R

⇒ α = R²α².t²/R² ⇒ α = 1/t²