A point moves in the xy-plane according to th | vedclass

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(A) The position vector of the point is given by $\vec{r} = x\hat{i} + y\hat{j} = a \sin(\omega t)\hat{i} + a(1 - \cos(\omega t))\hat{j}$.The velocity

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$\frac{\pi}{2}$

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(A) The position vector of the point is given by $\vec{r} = x\hat{i} + y\hat{j} = a \sin(\omega t)\hat{i} + a(1 - \cos(\omega t))\hat{j}$.The velocity vector $\vec{v}$ is the time derivative of the position vector:$\vec{v} = \frac{d\vec{r}}{dt} = a\omega \cos(\omega t)\hat{i} + a\omega \sin(\omega t)\hat{j}$.The acceleration vector $\vec{a}$ is the time derivative of the velocity vector:$\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2 \sin(\omega t)\hat{i} + a\omega^2 \cos(\omega t)\hat{j}$.To find the angle $	heta$ between $\vec{v}$ and $\vec{a}$,we use the dot product formula: $\cos 	heta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|}$.First,calculate the dot product $\vec{v} \cdot \vec{a}$:$\vec{v} \cdot \vec{a} = (a\omega \cos(\omega t))(-a\omega^2 \sin(\omega t)) + (a\omega \sin(\omega t))(a\omega^2 \cos(\omega t))$$\vec{v} \cdot \vec{a} = -a^2\omega^3 \cos(\omega t)\sin(\omega t) + a^2\omega^3 \sin(\omega t)\cos(\omega t) = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$,where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.

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