Simple calculation problems?

1. Jan 23, 2005

Mivz18

Problem: The consumption of natural gas by a company satisfies the empirical equation V=1.50t + 0.800 00(t squared), where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Assign proper units to the coefficients. Assume a month is equal to 30.0 days.

My work:

From the equation given, I took each part of the right and converted the months variable to seconds.

1.50(t months) * (t 30 days) * (t 24 hours) * (t 60 min) * (t 60 sec.)
-------------- ---------- ---------- --------- ----------
1 (t month) (t 1 day) (t 1 hr.) (t 1 min)

= 3.888E6(t sec.)

Then I repeated the conversion multiplication fractions with 0.008 00 and squaring the other numbers and reaching a value of 5.37477E9(t sec)^2

My new equation is then V= 3.888E6(t sec.) + 5.37477E9(t sec.)^2

dividing by a million to convert millions of cubic feet to cubic feet, I get:

V=3.889E6(t sec.) + 53747.7(t sec.)^2

However, the book gives an answer of 0.579t (ft cubed/sec.) + 1.19E-9t^2 (ft cubed/sec. squared)
Is not assigning units to the coefficients screwing me up or is there another problem with my calculations?

2. Jan 24, 2005

learningphysics

Remember that the given formula requires t (the time in months). When dealing with t you have to be careful... I'd do it like this:

let a=time in seconds.

a = t months * (30days/1 month) * (24 hours/1day)*(60min/1hr)*(60s/1min)
so a= 2.592E6 * t seconds

So now solve for t, and then plug back into the original equation.

So t=a/2.592E6 . Plug a/2.592E6 into the original equation. Think of it like this. You've got the time in seconds a, then getting the time in months t, in terms of a...

So you'll then get an equation in terms of a, where a is in seconds. Volume is still in millions of cubic feet.

To get from millions of cubic feet to cubic feet, you need to multiply by a million, not divide.