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sharp21 December 19th, 2007 06:43 PM

Need help with a problem...
 
Hey gang.
Ive been out of school for a number of years & am now preparing for a marine exam. The question has to do with fuel consumption.
But this one question has me stumped! I dont think it should be a difficult question, but im definately missing something. Here it is:

A vessel 800miles from port is forced to reduce its daily fuel consumption by 42tonnes per day, in order to arrive in port with 50tonnes remaining onboard. This will reduce her speed by 20%
If the fuel consumption measured in tonnes/hr is given by the expression (0.136+0.001V^3), where V is the speed in knots (basically mph) then calculate:
1. Reduced consumption per Day
2. Amount of fuel onboard at the time the speed was decreased.
3. Percentage decrease in the consumption for the latter part of voyage.

sharp21 December 19th, 2007 06:54 PM

Obviously you need to find the original speed first - V
Here is how that part of the problem is solved on the sample paper im woring from:

Let C = Normal consumption per hour at 'V' knots
Let C"=Reduced consumption per hour at '0.8V' (80% of original speed)

C=0.136+0.001v^3
C"=C-(42/24)

SO

C" - (42/24) = 0.0136 + 0.001v^3 (0.8v)^3
(this is where it all goes pear shaped on me...)

-42/24 = 0.001v^3 - 0.001 (0.512)

=0.001v^3 - 0.000512

v^3 = 42/24 (0.001 - 0.000512)

v^3 = 42/0.011712

V^3 = 3586.066

V = 15.3 kts

Answer = 15.3kts

What am i missing??
S.

sharp21 December 20th, 2007 06:08 PM

bump

skipjack December 21st, 2007 01:11 AM

Let C tonnes = Normal consumption per hour at 'V' knots
Let C" tonnes = Reduced consumption per hour at '0.8V' (80% of original speed)

C = 0.136 + 0.001V³
C" = C-(42/24)
C" = 0.136 + 0.001(0.8V)³

so 42/24 = C - C" = 0.001(1 - 0.8³)V³

so V³ = (42/24)/(0.001(1 - 0.512)) = 3586.06557...
so V = 15.306...
etc.

Does that help you enough?

sharp21 December 21st, 2007 04:35 PM

Close, but im still a lil confused on this part:

so 42/24 = C - C" = 0.001(1 - 0.8³)V³

I understand how you get 42/24=C-C" but not how the equation is derived...

C-C" would be:

0.136 + 0.001V³ - 0.136 + 0.001(0.8V)³
so the 0.136 & -0.136 would cancel, leaving:
0.001V³ + 0.001(0.8V)³
but I cant figure how to get it from there to:
0.001(1 - 0.8³)V³

Im not sure where the 1-o.8 comes from, or where the 2nd V³ goes...
S.

sharp21 December 21st, 2007 04:46 PM

So close...
Looking at what I just worked out, after cancelling the 0.136's

42/24= 0.001V³ + 0.001(0.8V)³

that should be a minus sign, as the whole of C" is being subtracted, so:


42/24= 0.001V³ - 0.001(0.8V)³

then if I cube the 0.8:

= 0.001V³ - 0.001(0.512) which is the same like the initial formula

but if I cube (0.8v)³ wouldnt it equal 0.512 x V³ ??
Where does that V go?

Sorry for being long winded... I know im just missing something simple, but im trying to understand, rather than just memorize...
S.

skipjack December 21st, 2007 08:59 PM

The V doesn't go! Cubing 0.8V gives 0.512 × V³,
so 0.001V³ - 0.001(0.8V)³ = 0.001V³ - 0.001(0.512V³).
Each term contains 0.001V³, so the expression can be factorized as (1 - 0.512)0.001V³.

Despite incorrectly dropping V³ and originally having an incorrect minus sign in front of 42/24,
you went on to write v^3 = 42/24 (0.001 - 0.000512), which was essentially correct,
although additional parentheses are needed for clarity: v^3 = 42/(24 (0.001 - 0.000512)).

sharp21 December 21st, 2007 09:49 PM

Ah! Eureka as they say..
Thanks for you help!
Now if I could just get my cheque book balanced...
S.


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