Quick test
Questions 4 to 6 refer to the following information:
2 On the other hand you can think about the question intuitively In which
case since the car covers 100 km every hour (a) it will take 1 hour to cover 100 km, (b) it will take twice this time to cover 200 km 2 hours and (c) 2.5 times as long to cover 250 km 2.5 hours.
A car moves at 100 km/hour, how long before it covers a distance of (a) 100 km, (b) 200 km and (c) 250 km?
Thinking about things intuitively helps to solve difficult problems and can often save time generally. Think about these in the same way:
What does it mean when the difference in speed between two objects is 5 m/s?
What does it mean when the average speed of two objects is 6 m/s?
Two buses move down a 100 km road at the same time moving at con- stant speeds. One bus gets to the end of the road 20 km ahead of the other bus.
What was the distance between the two buses after (a) the faster bus had covered 1/8th of the total distance and (b) the slower bus had covered 1/8th of the total distance? (a) If you think about this question intuitively then you can solve it pretty
quickly.
Since the two buses move at constant speeds and after 100 km the slower bus is lagging behind by 20 km it means that after 1/8th of 100 km the slower bus will be lagging behind by 1/8th of 20 km.
So the answer is 1/8th of 20 2.5 km.
On the other hand you can work it out using algebra (another very useful tool in many situations).
Let’s call the fast bus’s distance dA and the slower bus’s distance dB. It is clear from the question that for every 100 km A travels, B travels 80 km and lags behind by 20 km, therefore dB 0.8dA and the difference in their dis- tance at any time, dA dB 0.2dA.
Therefore when dA is 100/8 12.5, the distance between is 12.5 0.2 2.5 km.
(b) In this novel scenario, the algebraic solution is quicker
dA dB 0.2dA
(we need to fi nd dA)
12.5 0.8dA
dA 12.5/0.8 15.625
so, dA dB 15.625 12.5 3.125 km
Intuitively, if we say the speed of the faster and slower buses in the given time, T are 100 and 80, respectively, to cover 1/8th of the total distance (100/8, i.e. 12.5 km) it will take 12.5/80 0.15625T.
In the same time the faster bus would have covered the distance 0.15625 100 15.625 km. So the difference in distance covered is 15.625 12.5 3.125 km.
This question illustrates how having a good grasp of the concept of speed more than the formulae can help you resolve these questions.
I have a bucket with a tap at the bottom which empties the bucket at a constant rate of 20 cm3/s. I decide to fi ll my bucket with water from a tap fl owing at a constant rate of 30 cm3/s.
(a) If my bucket takes 5 minutes to fi ll up, what is the volume of my bucket?
This question is asking you to compare rates. You have the rate of fl ow into the bucket and the rate of fl ow out of the bucket. The actual rate at which the bucket is being fi lled is 30 cm3/s 20 cm3/s 10 cm3/s.
Therefore in 5 minutes (or 300 seconds) the total volume that would have communicated is 10 cm3/second 300 seconds 3000 cm3.
(b) If my bucket has a uniform cross-sectional area of 300 cm2 what is the height of my bucket?
This question now calls on your understanding of area and volume, in this case to get the height you need to divide the volume by the area.
Cross-sectional area height Volume Therefore:
Height Volume/cross-sectional area
Height of bucket 3000 cm3/300 cm2 10 cm
Bucket inflow tap 30 cm3/s
Bucket outflow tap 20 cm3/s
(c) I now have this exact same bucket but operating by different rules: the rate of fl ow from the tap at the bottom of the bucket is not constant but dependent on the height of water in the bucket. It is initially 20 cm3/s but increases linearly by 10 cm3/s for every 1 cm increase in the height of fl uid in the bucket. If I start fi lling the bucket at 30 cm3/s as before (i) what is the maximum height the fl uid will reach and (ii) how long will it take to get there?
Now the question has got a bit more complicated; it is asking you to compare two dynamic rates that are dependent on each other. This kind of question does not lend itself very easily to a formulaic approach which you may be used to. This is why it is important to have an intuitive grasp of the problem. (i) Initially the net rate is 10 cm3/s, as we have seen. But this changes very
quickly as the bucket fi lls up because the bucket begins to empty at a faster rate.
Since we know the way the bucket empties we can predict that it would begin to empty at 30 cm3/s when it reaches a height of 1 cm. At this stage there is a net rate of fl ow of 0 cm3/s. So it means this is the maximum height the fl uid would reach.
(ii) Finding the time it takes to reach the maximum height of 1 cm is a bit trickier because the net rate at which the bucket fi lls up is not constant. It fi lls up fast initially but slows down till it gets to zero. What we need is some sort of average rate we can apply. The clue is that the rate of exit from the bucket has a linear relationship to the height of fl uid in the bucket.
For simplicity let us take the average rate during that period as 25 cm3/s (halfway between 20 cm3/s and 30 cm3/s).
So all we have to fi nd out is how long will it take to get to the height of 1 cm at a net rate of (30 cm3/s 25 cm3/s) 5 cm3/s. To know this we need to divide the volume by the rate. The volume at 1 cm height is 300 cm3 (1 cm 300 cm2).
So the time taken to fi ll this volume will be (300 cm3/second)/(5cm3/second) 60 seconds.
30 cm3/s
Rate of inflow Rate of outflow
Average rate of approximate out flow 20 cm3/s 25 cm3/s Rate
Height of water in bucket 1 cm
Approximate rate of outflow
Concentration and other rates
Concentration is a concept you will be familiar with from chemistry but you can also think of it as a kind of rate as the amount of solute changes in a given way with the amount of solution. There a lot of interesting problems that concentrations can throw up.
A more complex problem: In addition to the first tap, I open up another tap which fills the bucket
simultaneously at 40cm3/s. Can you draw a graph of
(i) rate of fluid exit against time; (ii) height of the bucket against time?
Billy has two labelled jars, exactly the same in every respect, each con- taining liquid of equal volume. Jar A contains water and Jar B contains milk. Billy takes a teaspoon of milk from Jar B and puts it into Jar A, mixing uniformly. A teaspoon of this new mixture and is then put into Jar B. At the end of this process, which of the following is true?
1 The volume of milk in Jar A is greater than the volume of water in Jar B