Question

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. 4.5 m/s
• B. 5.0 m/s
• C. 5.7 m/s
• D. 6.2 m/s

Answer 1

The Correct Option is C

Calculation of Velocity (V) using given parameters 

We are given the following equation:

\[\operatorname{Tan} \theta = \frac{ \frac{dV}{dt} }{ \frac{V^2}{R} }\]

Now, substitute \( \theta = 30^{\circ} \), where \( \operatorname{Tan} 30^{\circ} = \frac{1}{\sqrt{3}} \), into the equation:

\[\operatorname{Tan} 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{ \frac{dV}{dt} }{ \frac{V^2}{R} }\]

This simplifies to:

\[dV = \frac{1}{\sqrt{3}} \times \frac{ V^2 }{ R }\]

Next, the acceleration \(a\) is given by:

\[a = 15 = \sqrt{\left( \frac{dV}{dt} \right)^2 + \left( \frac{V^2}{R} \right)^2}\]

Substitute the value of \(dV\) into the equation:

\[15 = \sqrt{\frac{4}{3} \left( \frac{ V^2 }{ R } \right)^2}\]

Square both sides to eliminate the square root:

\[15^2 = \frac{4}{3} \left( \frac{ V^2 }{ R } \right)^2\]

Simplify the equation:

\[225 = \frac{4}{3} \left( \frac{ V^2 }{ R } \right)^2\]

Now, isolate \( \frac{V^2}{R} \):

\[\left( \frac{ V^2 }{ R } \right)^2 = \frac{225 \times 3}{4} = 168.75\]

Now take the square root of both sides:

\[\frac{ V^2 }{ R } = \sqrt{168.75} = 12.99\]

Finally, solve for \( V^2 \) using the equation:

\[V^2 = \frac{15 \times \sqrt{3} \times 2.5}{2}\]

After simplifying:

\[V^2 = 32.476\]

The velocity \( V \) is:

\[V = \sqrt{32.476} = 5.7 \, \text{m/s}\]

Answer 2

Given: 
- Total acceleration $a = 15\, m/s^2$
- Radius of the circular path $R = 2.5\, m$

In circular motion, total acceleration is given by:
\(a = \sqrt{a_c^2 + a_t^2}\)
where:
 

• $a_c = \frac{v^2}{R}$ is the centripetal acceleration
• $a_t$ is the tangential acceleration

But since direction is not specified and only speed is asked, we assume motion is uniform (or tangential acceleration is zero):
So, \(a = a_c = \frac{v^2}{R}\)

Now solving for $v$:
\(v^2 = a \cdot R = 15 \cdot 2.5 = 37.5\)
\(\Rightarrow v = \sqrt{37.5} \approx \boxed{5.7\, m/s}\)

Final Answer:
\(\boxed{5.7\, m/s}\)