Question

A particle moves along a circle of radius $\left(\frac{20}{\pi}\right) m$ with constant tangential acceleration. If the velocity of the particle is $80\, m/s$ at the end of the second revolution after motion has begun, the tangential acceleration is

• A. 40 $ m/s^2$
• B. 640 $ \pi \, m/s^2$
• C. 160 $ \pi \, m/s^2$
• D. $40\, \pi \, m/s^2$

Answer 1

The Correct Option is A

To determine the tangential acceleration, let's analyze the motion step-by-step:

1. Given Data:
   
   • Radius of the circle: $r = \frac{20}{\pi} \, m$
   • Final velocity after the second revolution: $v = 80 \, m/s$
   • Initial velocity: $u = 0 \, m/s$ (since the motion starts from rest)
2. Circumference of the Circle:
   The circumference of the circle is given by:
   \[ C = 2 \pi r = 2 \pi \times \frac{20}{\pi} = 40 \, m \]
3. Distance Covered in Two Revolutions:
   The distance covered in two complete revolutions is:
   \[ s = 2 \times 40 = 80 \, m \]
4. Using the Equation of Motion:
   We use the equation of motion:
   \[ v^2 = u^2 + 2as \]
   Substituting the known values:
   \[ 80^2 = 0 + 2 \cdot a \cdot 80 \] \]
5. Solving for Tangential Acceleration:
   Rearrange the equation to solve for a:
   \[ 6400 = 160a \] \]
   \[ a = \frac{6400}{160} = 40 \, m/s^2 \] \]

Therefore, the tangential acceleration of the particle is 40 $ m/s^2$. This matches with the correct option provided.