If angular position of a particle is given by θ = t⁴ /4 + t². Find angular acceleration at t = 1s:

The correct option is 5rad/s².

To find the angular acceleration of the particle, we start from the given angular position θ as a function of time $t$:

$\theta =\frac{t^4}{4}+t^2$

First, we find the angular velocity $\omega$ by taking the first derivative of the angular position with respect to time:

$\omega =\frac{d\theta }{dt}=\frac{d}{dt}\left(\frac{t^4}{4}+t^2\right)$

Using the power rule for differentiation ($\frac{d}{dt}\left(t^n\right)=n{t}^{n-1}$):

$\omega =\frac{4t^3}{4}+2t=t^3+2t$

Next, we find the angular acceleration $\alpha$ by taking the derivative of the angular velocity with respect to time:

$\alpha =\frac{d\omega }{dt}=\frac{d}{dt}\left(t^3+2t\right)$

Applying the power rule again:

$\alpha =3t^2+2$

Now, we substitute $t=1s$ into the expression for angular acceleration to find its value at that instant:

$\alpha \left(1\right)=3{\left(1\right)}^2+2$
$\alpha \left(1\right)=3\left(1\right)+2=5{\text{rad/s}}^2$

Thus, the angular acceleration of the particle at $t=1s$ is 5rad/s².