Speed of the image (as seen by an observer in air) of the coin if the liquid surface is raised with speed $8\ m/s$

I know how to deal with when the object is moving but how to do when surface moves?

3 votes

Speed of the image (as seen by an observer in air) of the coin if the liquid surface is raised with speed $8\ m/s$

I know how to deal with when the object is moving but how to do when surface moves?

3 votes

Best answer

Suppose the observer is a fixed distance $H$ above the coin. Then the distance from the observer to the water surface is $H-h$ and the apparent distance of the coin below the surface is $\frac34 h$. So the apparent distance of the coin from the observer (measuring downwards) is $$y=(H-h)+\frac34h=H-\frac14h$$

The velocity of the water surface is $\frac{dh}{dt}=+8m/s$ (plus because $h$ is increasing) so the rate at which the image of the coin is moving downwards (in the direction of increasing y) is $$\frac{dy}{dt}=-\frac14\frac{dh}{dt}=-2m/s$$ The minus sign indicates that the image is moving upwards towards the observer, whereas y increases downwards.

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