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Physics · MOTION IN A PLANE
हिन्दीThe position vector of a particle moving in the \( x\text{-}y \) plane is given by \( \mathbf{r}(t) = (A\cos\omega t)\hat{i} + (A\sin\omega t)\hat{j} \), where \( A \) and \( \omega \) are positive constants. What is the angle between its velocity vector \( \mathbf{v}(t) \) and acceleration vector \( \mathbf{a}(t) \)?
Correct answer · Option 4
\( 90^\circ \)
Explanation
\( \mathbf{v} = -A\omega\sin\omega t \hat{i} + A\omega\cos\omega t \hat{j} \) and \( \mathbf{a} = -A\omega^2\cos\omega t \hat{i} - A\omega^2\sin\omega t \hat{j} \). Taking the dot product: \( \mathbf{v} \cdot \mathbf{a} = A^2\omega^3\sin\omega t\cos\omega t - A^2\omega^3\sin\omega t\cos\omega t = 0 \). Thus the angle is \( 90^\circ \).
