A particle starts from rest and its angular d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The angular displacement is given by $	heta = \frac{t^{2}}{20} + \frac{t}{5}$.Angular velocity $\omega$ is the rate of change of angular displace

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The angular displacement is given by $	heta = \frac{t^{2}}{20} + \frac{t}{5}$.Angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d	heta}{dt}$.Differentiating $	heta$ with respect to $t$:$\omega = \frac{d}{dt} \left( \frac{t^{2}}{20} + \frac{t}{5} ight) = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5}$.At $t = 4 \ s$,the angular velocity $k$ is:$k = \left( \frac{4}{10} + \frac{1}{5} ight) = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6 \ rad/s$.We need to find the value of $5k$:$5k = 5 	imes 0.6 = 3$.

$A$ particle starts from 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 $t = 4 \ s$ is $k$,then the value of $5k$ is

Similar Questions

Find the derivative of $f$ given by $f(x) = \sin^{-1} x$ assuming it exists.

Find the derivative: $\frac{d}{dx} \left( \frac{e^x}{1 + x^2} \right)$

The value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$ is:

If $f(x) = (x - x_0)g(x)$,where $g(x)$ is continuous at $x_0$,then $f'(x_0)$ is equal to

Find the derivative of $x^{-4}(3-4x^{-5})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module