Tap A and B

Classical Mechanics Level 3

Taps A A and B B together can fill a bucket in 6 minutes. Their rates (in which they each can fill the bucket) are different and accelerate as follows:

  • Tap A alone can fill the bucket in 8 minutes with initial speed 2 mL / s \SI[per-mode=symbol]{2}{\milli\liter\per\second} and acceleration 2 mL / s 2 . \SI[per-mode=symbol]{2}{\milli\liter\per\second\squared}.
  • Tap B alone can fill the bucket in x x minutes with initial speed 3 mL / s \SI[per-mode=symbol]{3}{\milli\liter\per\second} and acceleration n mL / s 2 \SI[per-mode=symbol]{n}{\milli\liter\per\second\squared} .

Find n n to two decimal places.


The answer is 1.54.

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Majed Kalaoun by Majed Kalaoun , 16, Canada

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1 solution

Steven Chase
Jun 24, 2017

Suppose V V is the volume of the bucket, r A r_A and r B r_B are the initial speeds, and a A a_A and a B a_B are the accelerations. The combined equation and the Tap A equation are below:

V = ( r A + r B ) ( 6 ∗ 60 ) + 1 2 ( a A + a B ) ( 6 ∗ 60 ) 2 V = ( r A ) ( 8 ∗ 60 ) + 1 2 ( a A ) ( 8 ∗ 60 ) 2 \large{V = (r_A + r_B)(6*60) + \frac{1}{2} (a_A + a_B) (6*60)^2 \\ V = (r_A)(8*60) + \frac{1}{2} (a_A) (8*60)^2}

Plugging in numbers gives:

V = ( 2 + 3 ) ( 6 ∗ 60 ) + 1 2 ( 2 + n ) ( 6 ∗ 60 ) 2 V = ( 2 ) ( 8 ∗ 60 ) + 1 2 ( 2 ) ( 8 ∗ 60 ) 2 \large{V = (2 + 3)(6*60) + \frac{1}{2} (2 + n) (6*60)^2 \\ V = (2)(8*60) + \frac{1}{2} (2) (8*60)^2}

We can solve for n n using the above equations, yielding n ≈ 1.54 n \approx 1.54 .

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