Solving Candle Burn Time Problem with Algebra

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Two candles, A and B, initially measuring 8cm and 14cm respectively, burn at different rates, with candle B burning twice as fast as candle A. When their lengths are equal, both candles measure 2cm, which occurs after 6cm of candle A's length has burned. Candle B will completely burn out after 7 time units, leaving candle A at 1cm in height. The algebraic approach simplifies the problem, confirming the solution found through a tabular method. This discussion highlights the effectiveness of algebra in solving real-world problems involving rates of change.
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Homework Statement



Two candles A and B are 8cm and 14cm long, respectively, and are lit at hte same time. If candle B burns twice as fast as candle A, how long are they when their lengths are the same. How long is candle A when candle B burns out?

Homework Equations



Not sure.

The Attempt at a Solution



I drew a table of length's for equal time width's. So when A is 7cm ling B will be 12cm long etc. I found that they would be 2cm a piece when they are the same length and that candle A would be 1cm long when candle B burns out. Is this correct?

More importantly how do I answer this question algebraically? The above seems to me like the dumb approach, but I just can't see how to model teh problem.

Any help would be much appreciated!
 
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Vuldoraq said:

Homework Statement



Two candles A and B are 8cm and 14cm long, respectively, and are lit at hte same time. If candle B burns twice as fast as candle A, how long are they when their lengths are the same. How long is candle A when candle B burns out?

Homework Equations



Not sure.

The Attempt at a Solution



I drew a table of length's for equal time width's. So when A is 7cm ling B will be 12cm long etc. I found that they would be 2cm a piece when they are the same length and that candle A would be 1cm long when candle B burns out. Is this correct?

More importantly how do I answer this question algebraically? The above seems to me like the dumb approach, but I just can't see how to model teh problem.

Any help would be much appreciated!

Your answers are right. Let LB = Length Burned of A

8 - LB = Height(a)

14 - 2*LB = Height(b)

When Heights are equal LB = 6, Height(a) = 2
When Height(b) = 0, T = 7, then Height(a) = 8 - 7 = 1
 
Thanks a million LowlyPion. I kept trying to use series sums etc, but I just couldn't get it to work. Your answer is much simpler!

Vuldoraq
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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