Topic: Uniform circular motion· Kinematics

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

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

circle radius R, total acceleration a at 30 degrees to radius

1. 5.0 m/s

2. 5.7 m/s

3. 6.2 m/s

4. 4.5 m/s

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 2 Practice test neet Save WhatsApp Report

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Answer: 2. 5.7 m/s

The total acceleration has a radial (centripetal) part $v^2/R$ and a tangential part. From the figure, $a$ makes $30^\circ$ with the line joining the particle to the centre, so the centripetal part is $a\cos30^\circ$.

$$\frac{v^2}{R}=a\cos30^\circ\Rightarrow v^2=2.5\times15\times0.866\approx32.5$$

$v\approx5.7$ m/s. Option (2).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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

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