The velocity vec{v} of a particle of mass m a | vedclass

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(A) The velocity vector is given by $\vec{v}(t) = A \cos(kt) \hat{i} - A \sin(kt) \hat{j}$.To find the force $\vec{F}$,we differentiate the velocity w

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(A) The velocity vector is given by $\vec{v}(t) = A \cos(kt) \hat{i} - A \sin(kt) \hat{j}$.To find the force $\vec{F}$,we differentiate the velocity with respect to time $t$:$\vec{F} = m \frac{d\vec{v}}{dt} = m \frac{d}{dt} [A \cos(kt) \hat{i} - A \sin(kt) \hat{j}]$$\vec{F} = m A [-k \sin(kt) \hat{i} - k \cos(kt) \hat{j}] = -mkA [\sin(kt) \hat{i} + \cos(kt) \hat{j}]$.Now,calculate the dot product of $\vec{F}$ and $\vec{v}$:$\vec{F} \cdot \vec{v} = (-mkA [\sin(kt) \hat{i} + \cos(kt) \hat{j}]) \cdot (A \cos(kt) \hat{i} - A \sin(kt) \hat{j})$$\vec{F} \cdot \vec{v} = -mkA^2 [\sin(kt)\cos(kt) - \cos(kt)\sin(kt)] = 0$.Since the dot product is zero,the angle between the force $\vec{F}$ and the velocity $\vec{v}$ is $90^{\circ}$.

The velocity $\vec{v}$ of a particle of mass $m$ acted upon by a constant force is given by $\vec{v}(t) = A[\cos(kt) \hat{i} - \sin(kt) \hat{j}]$. Then the angle between the force and the velocity of the particle is (Here $A$ and $k$ are constants). (in $^{\circ}$)

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