Rates of change question help (2 Viewers)

[eyeseeyou]

How do you do this?

On a factory production line a tap opens and closes to fill containers with liquids. 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

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

 

[davidgoes4wce]

Do you happen to have the answer for part b)? Just want to compare my answer

 

[eyeseeyou]

Do you happen to have the answer for part b)? Just want to compare my answer

I think it was 100 i can't remeber tbh

For this question, you had a diagram to illustrate the rate of change

 

[parad0xica]

Please post up the diagram.

 

[Nailgun]

Are you sure you wrote q1 right lel
maybe im being retarded but I get 6/5t - 6

edit: wait they may be the same thing i think

edit2: yea im retarded the expressions are equivalent

 

[Nailgun]

 

[eyeseeyou]

U got it wrong nailgun. It was 100 for one of the question iirc

 

[Shuuya]

U got it wrong nailgun. It was 100 for one of the question iirc

Can you confirm the answer please, I got t=105 secs for (b)

 

[leehuan]

You all interpreted the question wrong for b)

The question asks for how long must the tap REMAIN FULLY open such that the 120L container is filled without spillage. If we want to continue filling the container we refuse to consider the equation where the tap begins to get closed

Hence

6t/5=120
t=100

 

[Shuuya]

Oops

 

[eyeseeyou]

You all interpreted the question wrong for b)

The question asks for how long must the tap REMAIN FULLY open such that the 120L container is filled without spillage. If we want to continue filling the container we refuse to consider the equation where the tap begins to get closed

Hence

6t/5=120
t=100

lol basically leehuan was the only one who got it right

Great job leehuan

Anyways could you please show me your working out to this question, it would be awesome if you could

Thanks

 

[davidgoes4wce]

Can you confirm the answer please, I got t=105 secs for (b)

I got your answer as well Shuuya. t=105 seconds.

I've seen this question from one of the trials from somewhere (which I have stored on my hard drive and it has a solution of t=90 seconds which I believe is wrong as well)

I visualized it as a diagram.

 

[Shuuya]

lol basically leehuan was the only one who got it right

Great job leehuan

Anyways could you please show me your working out to this question, it would be awesome if you could

Thanks

leehuan showed his working out in the post above lol

Hence

6t/5=120
t=100

 

[davidgoes4wce]

Actually I reckon Nailgun might be right. I got 95 seconds as well .

 

[InteGrand]

I get 90 seconds.

 

[leehuan]

^Probably should not doubt that answer

 

[InteGrand]

We need to include a part in the R vs. t graph that slopes down to 0 (symmetrical to the initial increase). This is because we are told the rate goes down at a constant rate for the last 10 seconds. 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).

The rate is constant at 6/5 for the horizontal part. Letting the unknown required time be T, we'd need (6/5)*T = 108 then. Solving for T yields T = 90.

 

[Nailgun]

okay maybe we should all spell out how we interpreted the question ahah


since it only wants the time that its fully open we don't include the first 10 seconds while it is progressively open
so 6L will be filled in that time, and the fully open tap will be working at the rate of 6/5L/second, w
6 + 6/5t = 120
where t is the number of seconds since the tap became fully open (which is what I thought it wanted)

t=95

 

[InteGrand]

I think it is meant to be interpreted as:

Stage 1 (time 0 to 10): Tap's rate increases uniformly from 0 to 6/5 L/s.

Stage 2 (time 10 to 10+T): Tap's rate stays flat at 6/5 L/s. (This is the stage where it is fully open.)

Stage 3 (time 10+T to 20+T (i.e. (10+T) + 10)): Tap's flow rate decreases uniformly to 0 L/s.

The reason for this interpretation is due to the filling mechanism described in the question.

Then given this and the fact that the total volume needs to be 120 L, we are asked to find T.

(So the equation we end up with 6 + (6/5)*T + 6 = 120.)

 

[davidgoes4wce]

I think it is meant to be interpreted as:

Stage 1 (time 0 to 10): Tap's rate increases uniformly from 0 to 6/5 L/s.

Stage 2 (time 10 to 10+T): Tap's rate stays flat at 6/5 L/s. (This is the stage where it is fully open.)

Stage 3 (time 10+T to 20+T (i.e. (10+T) + 10)): Tap's flow rate decreases uniformly to 0 L/s.

The reason for this interpretation is due to the filling mechanism described in the question.

Then given this and the fact that the total volume needs to be 120 L, we are asked to find T.

(So the equation we end up with 6 + (6/5)*T + 6 = 120.)

I, initially thought like you, that had to be a constant rate in there somewhere, and it had to decrease back to 0.