To find the time when the angle between the velocity and acceleration vectors is $45^\circ$, we need to consider the components of acceleration in circular motion. Step 1: Identify the components of acceleration. In circular motion, the total acceleration $\vec{a}$ has two perpendicular components: • Tangential acceleration ($a_t$): This component is responsible for changing the speed of the body and is directed along the tangent to the circular path, in the same direction as the velocity vector $\vec{v}$. $$a_t = r\alpha$$ where $r$ is the radius of the circular path and $\alpha$ is the angular acceleration. • Centripetal acceleration ($a_c$): This component is responsible for changing the direction of the velocity and is directed towards the center of the circular path, perpendicular to the velocity vector. $$a_c = r\omega^2$$ where $\omega$ is the angular velocity. Step 2: Relate the angle to the acceleration components. The velocity vector $\vec{v}$ is tangential. The tangential acceleration $\vec{a_t}$ is parallel to $\vec{v}$. The centripetal acceleration $\vec{a_c}$ is perpendicular to $\vec{v}$. The angle $\theta$ between the total acceleration $\vec{a}$ and the velocity $\vec{v}$ (or tangential acceleration $\vec{a_t}$) can be found using trigonometry. In the right triangle formed by $\vec{a_t}$, $\vec{a_c}$, and $\vec{a}$, the angle $\theta$ satisfies: $$\tan \theta = \frac{a_c}{a_t}$$ Step 3: Express angular velocity in terms of time. The body starts from rest, so its initial angular velocity $\omega_0 = 0$. Given constant angular acceleration $\alpha = 4 \text{ rad/s}^2$. The angular velocity at time $t$ is: $$\omega = \omega_0 + \alpha t$$ $$\omega = 0 + (4 \text{ rad/s}^2)t$$ $$\omega = 4t \text{ rad/s}$$ Step 4: Substitute expressions for $a_t$, $a_c$, and $\omega$ into the tangent equation. $$\tan \theta = \frac{r\omega^2}{r\alpha}$$ $$\tan \theta = \frac{\omega^2}{\alpha}$$ Substitute $\omega = 4t$: $$\tan \theta = \frac{(4t)^2}{4}$$ $$\tan \theta = \frac{16t^2}{4}$$ $$\tan \theta = 4t^2$$ Step 5: Solve for time $t$. We are given that the angle $\theta = 45^\circ$. We know that $\tan 45^\circ = 1$. $$1 = 4t^2$$ $$t^2 = \frac{1}{4}$$ $$t = \sqrt{\frac{1}{4}}$$ $$t = \frac{1}{2} \text{ s}$$ $$t = 0.5 \text{ s}$$ The time at which the angle between velocity and acceleration is $45^\circ$ is: $$\boxed{\text{0.5 s}}$$ Send me the next one 📸