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In the given figure, a = 15 m/$s^2$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is:
A
5.7 m/s
B
6.2 m/s
C
4.5 m/s
D
5.0 m/s
Explanation
The radial component of the total acceleration equals v²/R.
Detailed Solution
Centripetal component: $a\cos30^\circ = a_c = \frac{v^2}{R}$
$\Rightarrow v^2 = aR\times\frac{\sqrt{3}}{2} = 15\times2.5\times\frac{\sqrt{3}}{2} = 32.47$
$\Rightarrow v \approx 5.7$ m/s
$\Rightarrow v^2 = aR\times\frac{\sqrt{3}}{2} = 15\times2.5\times\frac{\sqrt{3}}{2} = 32.47$
$\Rightarrow v \approx 5.7$ m/s