Rate of Flow question (1 Viewer)

[1729]

On a factory production line a tap opens and closes to fill containers with liquid. As the tap opens, the rate of flow increases for the first 10 seconds according to the relation R = 6t/50, where R is measured in L/sec. The rate of flow then remains constant until the tap begins to close. As the tap closes, the rate of flow decreases at a constant rate for 10 seconds, after which time the tap is fully closed.

(i) Show that, while the tap is fully open, the volume in the container at any time is given by V = 6/5 * (t-5)
(ii) For how many seconds must the tap remain fully open in order to exactly fill a 120L container with no spillage.

For (ii) I got 95 seconds but the answer is 90 seconds, any help appreciated

 

[InteGrand]

On a factory production line a tap opens and closes to fill containers with liquid. As the tap opens, the rate of flow increases for the first 10 seconds according to the relation R = 6t/50, where R is measured in L/sec. The rate of flow then remains constant until the tap begins to close. As the tap closes, the rate of flow decreases at a constant rate for 10 seconds, after which time the tap is fully closed.

(i) Show that, while the tap is fully open, the volume in the container at any time is given by V = 6/5 * (t-5)
(ii) For how many seconds must the tap remain fully open in order to exactly fill a 120L container with no spillage.

For (ii) I got 95 seconds but the answer is 90 seconds, any help appreciated

It's due to how you interpreted the question. Have a read of the discussion here (same question got asked and people appeared to misinterpret it): http://community.boredofstudies.org...sion-1/350318/rates-change-question-help.html .

 

[InteGrand]

"Fully open" means its flow rate is 6/5 L/sec, so we want the time for which the tap is operating at 6/5 L/sec.

 

[1729]

"Fully open" means its flow rate is 6/5 L/sec, so we want the time for which the tap is operating at 6/5 L/sec.

So using symmetry, the area under the decreasing part at the end will also be 6. Then the area under the constant part needs to be 108 (i.e. 120 – 6 – 6).

Thanks

I did this but I let V = 108 in V = 6/5 * (t-5) when I should've just found the area under the horizontal part.
Why does substituting V = 108 in V = 6/5 * (t-5) not work?

Edit: (btw, I got the question out of a HSC Trial)

 

[InteGrand]

Thanks

I did this but I let V = 108 in V = 6/5 * (t-5) when I should've just found the area under the horizontal part.
Why does substituting V = 108 in V = 6/5 * (t-5) not work?

Edit: (btw, I got the question out of a HSC Trial)

So in other words, V doesn't denote the area under the horizontal part only, it denotes the area under the horizontal part + the area under the initial triangular part. Hence we don't set V equal to 108, but rather to 114 (area of initial triangle plus 108).