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CGP EDU Academic Team
Published on: September 11, 2026
If the equation for the angular displacement of a particle moving on a circular path is given by ( θ ) = 2t 3 + 0.5, where θ is in radians and t in seconds, then find the angular velocity of the particle after 2 seconds from its start.
Text Solution
Verified by ExpertsThe correct answer is:
B
To find the angular velocity, we first need to determine the angular displacement function and then differentiate it with respect to time.
Step 1: The given angular displacement is \( \theta(t) = 2t^3 + 0.5 \).
Step 2: To find the angular velocity \( \omega \), we differentiate \( \theta(t) \) with respect to time \( t \):
\( \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 + 0.5) \).
Step 3: Calculating the derivative:
\( \omega(t) = 6t^2 \).
Step 4: We now evaluate this expression at \( t = 2 \) seconds:
\( \omega(2) = 6(2)^2 = 6 \times 4 = 24 \) radians/second.
Therefore, the angular velocity of the particle after 2 seconds is 24 radians/second.
Step 1: The given angular displacement is \( \theta(t) = 2t^3 + 0.5 \).
Step 2: To find the angular velocity \( \omega \), we differentiate \( \theta(t) \) with respect to time \( t \):
\( \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 + 0.5) \).
Step 3: Calculating the derivative:
\( \omega(t) = 6t^2 \).
Step 4: We now evaluate this expression at \( t = 2 \) seconds:
\( \omega(2) = 6(2)^2 = 6 \times 4 = 24 \) radians/second.
Therefore, the angular velocity of the particle after 2 seconds is 24 radians/second.
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