A man moves in the x-y plane along the path s | vedclass

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Question

(C) The average velocity vector $\vec{v}_{avg}$ from the starting point $P$ to any point $Q$ on the path is given by $\vec{v}_{avg} = \frac{\vec{r}_Q

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$C$

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(C) The average velocity vector $\vec{v}_{avg}$ from the starting point $P$ to any point $Q$ on the path is given by $\vec{v}_{avg} = \frac{\vec{r}_Q - \vec{r}_P}{t_Q - t_P}$,which is the direction of the displacement vector $\Delta \vec{r} = \vec{r}_Q - \vec{r}_P$.The instantaneous velocity vector $\vec{v}_{inst}$ at any point is tangent to the path at that point.For the average velocity vector to be in the same direction as the instantaneous velocity vector,the displacement vector $\vec{r}_Q - \vec{r}_P$ must be tangent to the path at point $Q$.Geometrically,this means the line segment connecting the starting point $P$ to point $Q$ must be tangent to the curve at point $Q$.Looking at the path,if we draw a line from $P$ to $C$,this line is tangent to the path at point $C$. Therefore,at point $C$,the average velocity vector is in the same direction as the instantaneous velocity vector.

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

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