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 December 19th, 2007, 07:43 PM #1 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 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.
 December 19th, 2007, 07:54 PM #2 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 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.
 December 20th, 2007, 07:08 PM #3 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 bump
 December 21st, 2007, 02:11 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,305 Thanks: 1976 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?
 December 21st, 2007, 05:35 PM #5 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 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.
 December 21st, 2007, 05:46 PM #6 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 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.
 December 21st, 2007, 09:59 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,305 Thanks: 1976 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)).
 December 21st, 2007, 10:49 PM #8 Newbie   Joined: Dec 2007 Posts: 8 Thanks: 0 Ah! Eureka as they say.. Thanks for you help! Now if I could just get my cheque book balanced... S.

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