32
𝑡𝑡
2
= 30
𝑘𝑘𝑘𝑘𝑘𝑘𝑡𝑡
2
=
𝑆𝑆
𝑣𝑣
2
→ 𝑆𝑆
=
𝑣𝑣
1
𝑡𝑡
1
𝑡𝑡
3
= ?
𝑡𝑡
3
=
𝑆𝑆
𝑣𝑣
1
+
𝑣𝑣
2
=
𝑣𝑣
1
𝑡𝑡
1
𝑣𝑣
1
+
𝑣𝑣
1
𝑡𝑡
1
𝑡𝑡
2
=
𝑡𝑡
1
𝑡𝑡
2
𝑡𝑡
1
+
𝑡𝑡
2
= 12
𝑘𝑘𝑘𝑘𝑘𝑘
Mathematical Method:
Let's define the work done with A and their work productivity with x and y respectively.
𝐴𝐴
𝑥𝑥
- How long the master completes all the work;
𝐴𝐴
𝑦𝑦
- How long the student completes his / her job;
𝐴𝐴
𝑥𝑥+𝑦𝑦
- How long the master and the student complete work together
1.
𝐴𝐴
𝑥𝑥
= 20
→ 𝑥𝑥
=
𝐴𝐴
20
2.
𝐴𝐴
𝑦𝑦
= 30
→ 𝑦𝑦
=
𝐴𝐴
30
3.
𝐴𝐴
𝑥𝑥+𝑦𝑦
=
𝐴𝐴
𝐴𝐴
20
+
𝐴𝐴
30
=
20∙30
20+30
= 12
Answer: It can be done in 12 days.
Case 2. There are 2 pipes in the pool. The first tube fills the empty pool in 10 hours. The
second tube empties in 15 hours. How long does it take to fill if bothtubesopens
when the pool is
empty?
Physical Method: In order to solve this problem, we use a physical formula that expresses the
dependence of the work on theenergy. As the size of the pool does not change, the work done to fill
or unload is equal. Because the pipes have different capacities, the time they go to work is different.
By such a comparison we resolve the matter.
Fig.1.
𝑡𝑡
1
= 10
𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑡𝑡
1
=
𝐴𝐴
𝑁𝑁
1
→ 𝐴𝐴
=
𝑁𝑁
1
𝑡𝑡
1
𝑡𝑡
2
= 15
𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑡𝑡
2
=
𝐴𝐴
𝑁𝑁
2
→ 𝐴𝐴
=
𝑁𝑁
2
𝑡𝑡
2
𝑡𝑡
3
= ?
𝑁𝑁
1
𝑡𝑡
1
=
𝑁𝑁
2
𝑡𝑡
2
→
𝑁𝑁
2
=
𝑁𝑁
1
𝑡𝑡
1
𝑡𝑡
2
𝑡𝑡
3
= (
𝐴𝐴
𝑁𝑁
1
− 𝑁𝑁
2
)
∗
=
𝑁𝑁
1
𝑡𝑡
1
𝑁𝑁
1
− 𝑁𝑁
1
𝑡𝑡
1
𝑡𝑡
2
=
𝑡𝑡
1
𝑡𝑡
2
𝑡𝑡
2
− 𝑡𝑡
1
=
10
∙
15
15
−
10 = 30
𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡
33
Note: The first tube fills the pool and the other empties it. When the two pipes are opened
together, the second pipe causes tofill slowly the pool, so the difference
of capacities of the two
pipes is taken.
Mathematical Method:
The overall capacitance of the pool does not change, so we find GCF (greatest common factor)
for filling and emptying.
10 = 2 × 5
15 = 3 × 5
GCF(10, 15) = 30 we can call it the capacity of the pool. The first pipe pumps 3 liters of water
an hour. The second tubepushes away 2 liters of water per hour.
30 = (3
−
2)
𝑡𝑡 → 𝑡𝑡
=
30
ℎ𝑠𝑠𝑘𝑘𝑜𝑜𝑠𝑠
isneeded to fill the pool.
Case 3. The radius of the front and rear wheels of a tractor is proportional to 5 and 7
respectively. If the car has been traveling on a 35
π
and the wheel has been rolling 20 times more
than the rear wheel, what is the radius of the front wheel?[2]
Physic method: We know from the physics that the linear velocities of the tractor wheels are
the same as that of the linear velocities in the transversal
transitions, and this is also equal to the
speed of the tractor. The distance traveled by the tractor is equal to the multiplication of wheel axis
length and the number of rolling of the wheel. Thus:
Do'stlaringiz bilan baham: