Question

The angular displacement of a particle is given by theta =t^3 + t^2 + t +1 then, its angular velocity at t=2 sec is …… rad s^(-1)

Answer 1

$\Large{\underline{\underline{\bf{Solution :}}}}$

We know that,

$\Large{\implies{\boxed{\boxed{\sf{\omega = \frac{d\theta}{dt}}}}}}$

Where,

$\theta$ = t³ + t² + t + 1

$\rule{150}{2}$

Now, Differentiating

$\sf{\omega = 3t^2 + 2t + 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (t = 2)} \\ \\ \sf{→\omega = 3(2)^2 + 2(2) + 1} \\ \\ \sf{→3(4) + 4 + 1} \\ \\ \sf{→12 + 5} \\ \\ \sf{→17 }$

$\Large{\implies{\boxed{\boxed{\sf{17 \: rad \: s^{-1}}}}}}$

Answer 2

ω=  

dt

dθ

​  

=3t  

2

+2t+1

at, t=2sec,

ω=12+4+1

=17 rad/s