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[dsheller]

Anyone want to lend a hand on this?

A water trough is 10 m long and a cross section has the shape of an isocelese trapezoid that has a base of 30 cm and a height of 50 cm and is 80 cm wide at the top. If it is being filled with water at a rate of 0.2 m^3 / minute how fast is it rising when the water is 30 cm deep...

I'm not sure what i'm doing on this one. Any help would be great.

[zaza]

Short edge of trapezoid = 30cm.
Long edge = 80cm.

Hence each right-angle triangle on the sides has base 25cm, because top = 25+30+25= 80cm.

Height of triangle = 50cm. When water is 30cm deep, this is 3/5 of the height of the trough. Similar triangles mean that the watery width of the triangle is thus 15cm. (Check if you like - the angle at the top is arctan(50/25) which is the same no matter how deep the water is).

So the horizontal width of water is 15+30+15 = 60cm. Hence the surface area of the water is 0.6m x 10m.

Not to finish it off for you... ...you know the surface area and you know what volume comes in in 1min, so you can work out the depth arriving in one min.
Which is how fast it is rising.

zaza

[dsheller]

Thanks! I tried something similar to that at one point but evidently messed up throughout the process.

[Rassis]

There is very little left to be said. Nonetheless, I would like to add a general expression for the instantaneous upward velocity “V” as a function of the elapsed time “t” and as a function of the water level “h”.

V = 3 / (20.t – 45)
V = 3 / (45 + 150.h)