Question

A particle starts form rest and its angular displacement (in radians) is given by \(\theta=\frac{t^2}{20}+\frac{t}{5}\). If the angular velocity at the end of t = 4 is k, then the value of 5k is

• A. 0.6
• B. 5
• C. 5k
• D. 3

Answer 1

The Correct Option is D

The angular displacement is given by the equation: \[ \theta = \frac{t^2}{20} + \frac{t}{5} \] To find the angular velocity \( \omega \), we need to take the first derivative of the angular displacement with respect to time \( t \): \[ \omega = \frac{d\theta}{dt} \] Differentiating \( \theta = \frac{t^2}{20} + \frac{t}{5} \) with respect to \( t \): \[ \omega = \frac{d}{dt} \left( \frac{t^2}{20} + \frac{t}{5} \right) \] The derivative of \( \frac{t^2}{20} \) is: \[ \frac{d}{dt} \left( \frac{t^2}{20} \right) = \frac{2t}{20} = \frac{t}{10} \] The derivative of \( \frac{t}{5} \) is: \[ \frac{d}{dt} \left( \frac{t}{5} \right) = \frac{1}{5} \] Therefore, the angular velocity is: \[ \omega = \frac{t}{10} + \frac{1}{5} \] Now, at \( t = 4 \), the angular velocity is: \[ \omega = \frac{4}{10} + \frac{1}{5} = \frac{2}{5} + \frac{1}{5} = \frac{3}{5} \] We are told that the angular velocity at \( t = 4 \) is \( k \), so \( k = \frac{3}{5} \). Therefore, the value of \( 5k \) is: \[ 5k = 5 \times \frac{3}{5} = 3 \] Thus, the correct answer is \({3} \), which corresponds to option (D).

Answer 2

Given the angular displacement \( \theta = \frac{t^2}{20} + \frac{t}{5} \), the angular velocity \( \omega \) is the derivative of \( \theta \) with respect to time \( t \).

Therefore, \( \omega = \frac{d\theta}{dt} = \frac{d}{dt} \left( \frac{t^2}{20} + \frac{t}{5} \right) \).

Calculating the derivative:

\( \omega = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5} \)

At \( t = 4 \), the angular velocity \( k \) is:

\( k = \omega(4) = \frac{4}{10} + \frac{1}{5} = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6 \)

We need to find the value of \( 5k \):

\( 5k = 5 \times 0.6 = 3 \)

Therefore, the value of 5k is 3.