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Mathematics Secondary 2. Chapter 1 Congruence and Similarity. Page 4

16. On a map the distance between two places A and B is 15 cm and the area of the Central Business District is 8 cm2. If the scale of the map is 1 : 80000,
(a) Find the actual distance of AB in km,
(b) Find the actual area of the Central Business District in km2.

17. The plans of a building are drawn to a scale of 1 : 150.
(a) Find the actual length, in metres, represented by 42 cm on the plan.
(b) The actual length of the diagonal of a hall is 33 m. Find its lenth, in cm, on the plan.
(c) The area of a meeting room is 450 m2. Find its area, in cm2, on the plan.

18. The diagram shows a triangle ABC where AB = 4 km, AC = 3 km and BC = 5 km. If  triangle ABC is drawn on a map of scale 1 : 50000, find
(a) the length of AB, in cm, on the map,
(b) the area of triangle ABC, in cm2, on the map.

19. A football stadium is represented by a scale of 1 cm to 8 m on paper.
(a) Find the actual length of the field if the length on the drawing is 28 cm.
(b) The actual width of the stadium is 120 m. Find its width on the diagram.
(c) If the area of the seating gallery is 960 m2. Find its area on the drawing.

20. The area scale of a map is 1 : 16000000. The length of two places A and B on the map is 8 cm and the area of a private housing estate is 12 cm2.
(a) Find the linear scale of the map in the form 1 : n.
(b) Find the actual distance of AB in metres.
(c) What is the area of the private housing estate in hectares?

Mathematics Secondary 2. Chapter 1 Congruence and Similarity. Page 3

11. Given that ABC triangle is similar to APQ triangle, calculate the values of x and y.

12. APQ triangle is similar to ABC triangle and AP : PB = 5 : 3. If PQ = 8 cm and QC = 6 cm, calculate the length
(a) BC,
(b) AQ.

13. A model of an apartment block is made to a scale of 1 : 50.
(a) Find the actual height of the apartment bloack if its height on the model is 42 cm.
(b) If the area of the hall of the apartment is 34 m2, find the area of the hall on the model.
(c) If the area of a unit of the apartment is 1200 cm2 on the model, find its actual area in m2.

14. On a map drawn on a scale of 2 cm to represent 300 m, what length on the map will represent a road 2.4 km long? A railway track on the map has a length of 14. 5 cm. Find its actual length in km.

15. A model of a building is made to a scale of 1 cm to 6 m.
(a) How tall is the model if the actual height of the building is 260 m?
(b) The base area of the building is 5400 m2. Find the base area of the building in the model.

Need more practise, go to number 16

Mathematics Secondary 2. Chapter 1 Congruence and Similarity. Page 2

6. PQRS is similar to ABCD
(a) Name three pairs of corresponding sides.
(b) Find the values of x and y.

(a) PQ = AB , QR = BC , CD = RS
(b) x = 102 , y = 70

7. The figure shows two similar cones.
(a) find the values of x.
(b) What is the ratio of the base circumference of the smaller cone to that of the big cone?

^h1/_h2 = ^r1/_r2
^x/₁₀ = ³/₅
x = ^{3 x 10}/₅
x = ³⁰/₅
x = 6
(b) to calculate the ratio: 2phi . r1 : 2phi . r2
eliminate (2phi) in every side.
r1 : r2
3 : 5

8. In the figure, BPC is parallel to DQE.
(a) Name three pairs of similar triangles.
(b) If AC = 6 cm, CE = 8 cm, CP = 5 cm, BP = 2 cm and BD = 10 cm, find the values of x, y and z.

9. ABC Triangle is similar to PRQ triangle. Find the values of x and y.

10. Given that PQR triangle is similar to XYZ triangle, calculate the values of x and y.

Need more practise, go to number 11

Secondary 2 Mathematics. Chapter 2 Direct and Inverse Proportion page 3

11. A local radio station has been given the exclusive rights to promote a concert in the city’s civic arena, which seats 22 000 persons. The commission $ C for the radio station is $ 5 000 plus $ 0.50 for each of the n tickets sold for the concert.
(a) Write down the formula connecting C and n.
(b) Use the formula to calculate
 (i) the commission for the radio station when 15 000 tickets are sold,
 (ii)the number of tickets sold when the commission for the radio station is $ 11 250,
 (iii)the maximum commission for the radio station.

(a)C = 5 000 + 0.5 n
(b) (i) C = 5 000 + 0.5 n
C = 5 000 + 0.5 (15 000)
C = 5 000 + 7500
C = 12 500
(ii) C = 5 000 + 0.5 n
11 250 = 5 000 + 0.5 (15 000)
0.5 n = 6 250
n = 12 500
(iii) C = 5 000 + 0.5 n
C = 5 000 + 0.5 (22 000)
C = 5 000 + 11 000
C = 16 000

12. (a) If a is directly proportional to the cube of b and that a = 54 when b = 3, find the equation relating a and b.
      (b) using the equation in (a), find the value of
            (i) a when b = 5, (ii) b when a = 128.

(a)
a = k . b³
54 = k . 3³
k = 54/27
k = 2
a = 2 . b³

(b)(i)
a = 2 . b³
a = 2 . 5³
a = 2 . 125
a = 250
(b)(ii)
a = 2 . b³
128 = 2 . b³
b³ = 128/2
b³ = 64
b = 4

13. In each of following, find the unknown variable without finding the proportionality constant.
(a) If h is directly proportional to the square of k and h = 8 when k = 3, find h when x = 6.
(b) If m is directly proportional to p^1/2 and m = 12 when p = 16, find p when m =18.

(a)
h = k²
h₁/h₂ = k₁²/k₁²
8/h₂ = 3²/6²
8/h₂ = 9/36
h₂ = 32
(b)
m = n^1/2
m₁/m₂ = n₁^1/2/n₂^1/2
12/18 = 4/n₂^1/2
n₂^1/2 = 6
n = 36

14. If the height of a cylinder is fixed, its volume is directly proportional to the square of its radius. When the radius of the cylinder is 2 cm, its volume is 81 cm³. Find the radius of the cylinder if the volume is 506.25 cm³.

V = k . r²
81 = k . 2²
k = 81/4
k = 20.25
so, V = 20.25 . r²
If the volume is 506.25, find the radius
V = 20.25 . r²
506.25 = 20.25 . r²
r² = 25
r = 5

15. Given that a²is directly proportional to b³ and a = 8(3)^1/2whenb = 4,
(a) find the equation connecting a and b,
(b) find the value of a when b = 3.

(a)
a²= k. b³
[8(3)^1/2]² = k . 4³
k = 192/64
k = 3
so, a²= 3 . b³
(a)
If b is 3, find a,
a²= 3 . b³
a²= 3 . 3³
a²= 81
a = 9

Secondary 2 Mathematics. Chapter 2 Direct and Inverse Proportion page 2

6.Two quantities P and Q are related by the formula  P = A - ^B/_Q², where A and B are constants. Given that P = 1 when Q = 2 and P = 6 when Q = 3,
(a) write down two equations in A and B,
(b) solve these equations to find the value of A and the value of B,
(c) find the positive value of Q when P = 7 ³/₄

(a) we know the equation is
Given that P = 1 when Q = 2 and P = 6 when Q = 3,
P = A - B/Q2
A = P + B/Q2
B/Q2
A = 1 + B/22
A = 1 + B/4
A = 4/4 + B/4
4A = 4 + B
4A - B = 4
and
P = A - B/Q2
A = P + B/Q2
A = 6 + B/32
A = 6 + B/9
A = 54/9 + B/9
9A = 54 + B
9A - B = 54
(b)
4A - B = 4-----first equation
9A - B = 54-----second equation
Subtract equation I and equation II, so we get
A = 10 and B = 36.
(c)
value of Q when P = 7 3/4
10 = 73/4 + 36/Q2
40/4 - 31/4 = 36/Q2
Q2 = 36 : 9/4
Q2 = 16
Q = 4

7. The intensity of illumination, I lumens/m², at a point on a screen is inversely proportional to the square of the distance, d m, of the light source from the point. Given that d = 2.5 m when I = 0.8 lumens/m², find
(a)  I, when d = 1.25 m, 

(b)  d, when I = 0.05 lumens/m².

Given that d = 2.5 m when, I = 0.8 lumens/m², find I = ^k/_d²
0.8 = ^k/_2.5²
k = 0.8 x 2.5² = 5
(a) I = ⁵/_1.25²
I = 3.2
(b) 0.05 = ⁵/_d²
d² = 5 / 0.05
d = 100^1/2 = 10

8. Water flows from a container such that the depth of water x cm at any instant, is inversely proportional to the square root of the time t s, for which the water has been flowing. After 25 s, the depth is 450 cm. Calculate the depth after

(a)  100 s,
(b)  3 min 45 s.

Given that After 25 s, the depth is 450 cm. x = ^k/_t^1/2
450 = ^k/_25^1/2
k = 450 x 5² = 2250
(a) x = ²²⁵⁰/_t^1/2
if t = 100 s,
x = ²²⁵⁰/_100^1/2
x = 2250 / 10 = 225 cm
(b) x = ²²⁵⁰/_t^1/2
if t = 225 s,
x = ²²⁵⁰/_225^1/2
x = 2250 / 15 = 150 cm

9. The resistance R, to the motion of a car is directly proportional to the speed v of the car. Given that the resistance is 2 688 newtons when the speed is 16 m/s, find

(a)  the resistance when the speed is 28 m/s,

(b)  the speed when the resistance is 16 800 newtons.

Given that the resistance is 2 688 newtons when the speed is 16 m/s. R = k.v
2688 = k.16
k = 2688/16 = 168
(a) if the velocity is 28 m/s,
R = 168.v
R = 168 x 28
R = 4704 Newton
(b) if the resistance is 16,800 N,
R = 168 x v
16800 = 168.v
v = 100 m/s

10.The total cost, $ c, of owning and operating a car is given by the formula c = a + bx, where x is the distance driven in km and a and b are constants. When the car is driven 2 500 km during it lifetime, the total cost is $ 19 000 and when the car is driven 6 000 km during its lifetime, the total cost is $ 20 400.
(a) Write down two equations in a and b.
(b) Solve these equations to find the value of a and of b.
(c) Find the total cost, if the car is driven a total distance of
(i) 50 000 km, (ii) 100 000 km.

(a)
c = a + bx
first equation
19000 = a + b(2500)
second equation
20400 = a + b(6000)
(b)
Eliminate first and second equation
19000 = a + b(2500)
20400 = a + b(6000)
__________________________ _
a = 18000 ; b = 2/5 = 0.4
(c)
The cost if the distance is 50000 km
c = 18000 + 0.4x
c = 18000 + 0.4 x 50000
c = 18000 + 20000
c = $38 000
The cost if the distance is 100 000 km
c = 18000 + 0.4x
c = 18000 + 0.4 x 100 000
c = 18000 + 40000
c = $58 000

go back to number 1-5

Secondary 2 Mathematics. Chapter 2 Direct and Inverse Proportion page 1

Chapter 2 Direct and Inverse Proportion


1. If q is inversely proportional to p, and q = 120 when p = 2, form an equation connecting p and q and calculate q when p = 5.

q = ^k/_p
120 = ^k/₂
k = 120 x 2 = 240
if p = 5 , so
q = ^k/_p
q = ²⁴⁰/₅
q = 48

2.         If y is inversely proportional to (2x + 1) and y = 5 when x = 3, find
(a)  y when x = 17,                  (b)  x when y = 7.

y = ^k/_{(2x + 1)}
5 = ^k/_[2(3)+1]
5 = ^k/₇
k = 5 x 7 = 35
(i) if x = 17 , so
y = ³⁵/_{[2(17) + 1]}
y = ³⁵/₃₅
y = 1
(ii) if y = 7, so
y = ³⁵/_{[2x + 1]}
7 = ³⁵/_{[2x + 1]}
35 = 7 x (2x + 1)
35 = 14x + 7
14x = 35 - 7
x = 28 / 14 = 2

3. If y is inversely proportional to the square of (3x + 2) and y = 4 when x = ²/₃  , find
(a)  y when x = 11 ¹/₃ , (b)  x when y = 16.

k = 64
(i) if x = 11¹/₃, so
y = 64 / [3(11¹/₃) + 2]²
y = 64 / 36²
y = 64 / 1296
y= 4/81
(ii) when y = 16, so
y = k / (3x + 2)²
16 = 64 / (3x + 2)²
(3x + 2)² = 64 / 16
(3x + 2) = 4^1/2
3x = 2 - 2
x = 0 / 3 = 0
or 3x = -2 - 2
x = -4 / 3 = -1¹/₃

4. Given that p is directly proportional to (2q + 1)^1/2  and p = 63 when q = 24, find
(a)  p when q = 12, (b)  q when p = 27.

p = k x (2q + 1)^1/2
63 = k x [2(24)+1]^1/2
63 = k x 49^1/2
k = 63 / 7 = 9
(i) if q = 12, so
p = 9 x (2(12) + 1)^1/2
p = 9 x 5
p = 45
(ii) when p = 27, so
p = k x (2q + 1)^1/2
27 = 9 x (2q + 1)^1/2
(2q + 1)^1/2 = 27 / 9
(2q + 1)^1/2 = 3
(2q + 1) = 3²
2q = 9 - 1
q = 8/2
q = 4

5. Given that d is directly proportional to the square root of t, copy and complete the table below.

The equation is d = k x t^1/2
8 = k x 4^1/2
k = 8 / 2
k = 4
(i) if t = 9, so we can calculate distance from that equation
d = k x t^1/2
d = 4 x 9^1/2
d = 4 x 3 d = 12
(ii) if d = 20, so the time is
d = k x t^1/2
20 = 4 x t^1/2
t^1/2= 20/4
t= 5²
t = 25
(iii) if t = 2¹/₄, so we can calculate distance from that equation
d = k x t^1/2
d = 4 x (2¹/₄)^1/2
d = 4 x (3/2)
d = 6

Need more practise, go to number 6

Mathematics Secondary 2. Chapter 1 Congruence and Similarity. Page 1

Chapter 1 Congruence and Similarity.
 
1.  ∆PQR is similar to ∆ABC. Given that PQ = 5 cm, QR = 4 cm and AB = 8 cm, calculate the length of BC.

^PQ/_QR = ^AB/_BC
⁵/₄ = ⁸/_BC
BC = ^{8 x 4}/₅
BC = ³²/₅=6²/₅

2.  ABCD is a trapezium where AB is parallel to DC. Name a pair of similar triangles. If
     AM = 5 cm, MC = 7 cm, BM = 6 cm, AB = 8 cm, DC = x cm and MD = y cm, find the
    values of x and y.

^DC/_AB = ^MC/_AM
^x/₈ = ⁷/₅
x = ^{8 x 7}/₅
x = ⁵⁶/₅=11¹/₅
^MD/_BM = ^MC/_AM
^y/₆ = ⁷/₅
y = ^{6 x 7}/₅
y = ⁴²/₅=8²/₅

3. State, with reasons, whether ∆ABC is similar to ∆PQR and find the value of x.

 

angle of P = 45, angle of C= 35, so ∆ABC is similar to ∆PQR
^QR/_BC = ^QP/_BA
^x/₇ = ⁶/₄
x = ^{6 x 7}/₄
x = ⁴²/₄=10¹/₂

4.  ∆OAB is similar to ∆OPQ.

(a) Explain clearly why AB is parallel to QP.

(b) If OA = 5 cm, OB = 6 cm, OQ = 8 cm, QP = 5.5 cm, OP = x cm and AB = y cm, find the values of x and y.

(a). The angle of Q = angle of B and they form alternate angles, AB // QP.
(b). ^OP/_OA = ^OQ/_OB
x = ^{(OQ x OA)}/_OB
x = ^{8 x 5}/₆=⁴⁰/₆
= 6²/₃
^AB/_QP = ^OB/_OQ
^y/_5.5 = ⁶/₈
y = ^{6 x 5.5}/₈
y = ³³/₈=4¹/₈

5. In Figure, ∆ABC is similar to ∆PQR. Given that AB = 8 cm, AC = 10 cm, PQ = 12 cm, QR = 9cm, BC = x cm and PR = y cm, calculate the values of x and y.

^BC/_QR = ^AB/_QP
^x/₉ = ⁸/₁₂
x = ^{8 x 9}/₁₂
x = ⁷²/₁₂=6
^RP/_CA = ^QP/_AB
^y/₁₀ = ¹²/₈
y = ^{12 x 10}/₈
y = 15

Need more practise, go to number 6

Pressure (In Solid and In Liquid) Worksheet and Answer For Junior High School

1. Calculate the pressure on a woman’s foot when a woman who weighs 500N steps on her foot with her heels which have total area of 0.5 m² with all her weight.

P = ^F/_A
P = ^{500 N}/_{0.5 m²}
P = 1000 Pa

2.Calculate the pressure exerted on the floor when an elephant who weighs 6000 N stands on one foot which has an area of 20 m²

Known:w = 6000 N
A = 20 m²
Asked: P?
Answer: P = ^w/_A
P = ^{6000 N}/_{20 m²}
P = 300 Pa

3. Calculate the pressure exerted on the heel of a girl’s foot if the girl weighs 500 N and she lands on one heel which has an area of 16 cm².

Known:w = 500 N
A = 16 cm² = 16 x 10^-4
Asked: P?
Answer: P = ^w/_A
P = ^{500 N}/16 x 10^-4
P = 312,500 Pa

4. A woman walking in high heals can damage a hardwood floor by making small dimplesin the floor since her weight is concentrated on such a small area (the tip of thehigh heal). If the woman weighs 500 N and the tip of the high heal is 1/15 m². what is the pressure exerted on the floor by her high heal?

Known:w = 500 N
A = 1/15 m²
Asked: P?
Answer: P = ^w/_A
P = ^{500 N}/(1/15)
P = 7500 Pa

5. A 75-kg solid cylinder, 2.5 m long and with an end radius of 5.0 cm, stands on one end. How much pressure does it exert?

6.    A barber raises his customer’s chair by applying a force of 150N to a hydraulic piston of area 0.01 m2. If the chair is attached to a piston of area 0.1 m2, how massive a customer can the chair raise? Assume the chair itself has a mass of 5 kg.

7.    A hydraulic system is used to lift a 2000-kg vehicle in an auto garage. If the vehicle sits on a piston of area 0.5 square meter, and a force is applied to a piston of area 0.03 square meters, what is the minimum force that must be applied to lift the vehicle?

8.    A 10 cm3 block of paraffin (a type of wax) weighs 0. 085 N. It is carefully submerged in a container of gasoline. One cm3 of gasoline weighs 0.0069 N.
a. What is the weight of the gasoline displaced by the paraffin?
b. Will the block of paraffin sink or float in the gasoline?

9.     A 70kg ancient statue lies at the bottom of the sea. Its volume is 3 x 104 cm3. How much force is needed to lift it?

10. When a crown of mass 14.7 is submerged in water, an accurate scale reads only 13.4 kg. Is  the crown made of gold? (ρgold = 19,300kg/m3)

Turning effect (Lever, Pulley, and Inclined Plane) Worksheet and Answer For Junior High School

1.First class lever has a 4900N weight located 4 meters from the fulcrum. If the force is applied 8 meters from the fulcrum, how much force is needed?

Known:w = 4900 N
l_A = 4 m (distance beetwen weight and pivot)
E_A = 8 m (distance beetwen force and pivot)
Asked: F?
Answer: w . l_A = F . E_A 4900 N . 4 m = F . 8 m
F = ^{4900 N . 4 m} / _{8 m}
F = 2450 Newton

2.Second class lever has a 19600N weight located 4 meters from the fulcrum.  If the lever is 10 meters long, how much force must be applied at the other end to lift the weight?

Known:w = 19600 N
l_A = 4 m (distance beetwen weight and pivot)
For second class lever:
Length of lever = E_A = 10 m (distance beetwen force and pivot)
Asked: F?
Answer: w . l_A = F . E_A 19600 N . 4 m = F . 10 m
F = ^{4900 N . 4 m} / _{8 m}
F = 7840 Newton

3. A third class lever is 4.8 meters long. The fulcrum is at one end. The force is applied 3.0 meters from the fulcrum.  How much force will it take to lift a weight of 160 N?

Known:w = 160 N
For third class lever:
Length of lever = l_A = 4.8 m (distance beetwen weight and pivot)
E_A = 3 m (distance beetwen force and pivot)
Asked: F?
Answer: w . l_A = F . E_A
160 N . 4.8 m = F . 3 m
F = ^{160 N . 4.8 m} / _{3 m}
F = 256 Newton

4. Two boys take a 12 m long board and rest it on a large rock to make a teeter-totter. If the 50 kg boy sits x m from the fulcrum, how far must the 70 kg boy sits from the fulcrum if they balance? (hint: use the 50 kg boy as output/object)

Known:
Length of lever = 12 m
the 50 kg boy as output/object)
w = 50 x 10 = 500 N
F = 70 x 10 = 700 N
l_A = x m (distance beetwen weight and pivot)
E_A = (12 - x) m (distance beetwen force and pivot)
Asked: E_A?
Answer: w . l_A = F . E_A
500 N . x m = 700 N . (12 - x) m
500x = 8400 - 700x
700x + 500x = 8400
1200x = 8400
x = 8400/1200 x = 7 m So, E_A = (12 - x) m (distance beetwen force and pivot)
= 12- 7
= 5 m

5. The man is lifting a box, it has mass 400 kg, how much force does the man need to apply to lift box?

Known:
mass of box = 40 kg, so weight = 40 x 10 = 400 N
Asked: F?
Answer: The mechanical advantages of pulley on that picture is 2 ( based on numbers of pulleys)
MA = W/F
2 = 400 / F
F = 400 / 2
F = 200 N

6. What is the MA of this ramp? Show your work.
a). Calculate the MA
b). Calculate the force do we need to apply to the box

Known:
mass of box = 100 kg, so weight = 100 x 10 = 1000 N
Base of inclined plane = 12 m
Slope of Inclined plane = 15 m
Asked: a). MA?
b). F?
Answer:
a). MA = s/h
We find the height using pythagorean theorm; a² + b² = c²
h²=s² - b²
= 15² - 12²
= 225 - 144
= 81
h = 81^1/2
h = 9 m,
MA = 15 / 9
= 1.67
b).MA = w/F
15/9 = 1000 / F
F = 600 N

Kinetic Energy and Potential Energy Worksheet and Answer For Junior High School

1. You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has ______________________ energy. Calculate it.

Kinetic
KE = ¹/₂ m v²
KE = ¹/₂ x 2.1 kg x (30 m/s)²
KE = ¹/₂ x 2.1 kg x 900 m²/s²
KE = 945 Joule

2. A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 N. The carriage has ____________ energy. Calculate it.

Gravitational Potential Energy
PE = m.g.h
note: weight = m. g =12 N
so put the weight directly to the equation below
PE = 12 N . 21 m
PE = 252 N.m
PE = 252 Joule

3. A car is traveling with a velocity of 40 m/s and has a mass of 1120 kg. The car has ___________energy. Calculate it.

Kinetic KE = ¹/₂ m v²
KE = ¹/₂ x 1120 kg x (40 m/s)²
KE = ¹/₂ x 1120 kg x 1600 m²/s²
KE = 896,000 Joule

4. A cinder block is sitting on a platform 20 m high. It weighs 79 N. The block has _____________ energy. Calculate it.

Gravitational Potential Energy
PE = m.g.h
note: weight = m. g = 79 N
so put the weight directly to the equation below
PE = 79 N . 20 m
PE = 1,580 N.m
PE = 1,580 Joule

5. There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________ energy. Calculate it.

Gravitational Potential Energy
PE = m.g.h
note: weight = m. g = 190 N
so put the weight directly to the equation below
PE = 190 N . 45 m
PE = 8,550 N.m
PE = 8,550 Joule

6. A roller coaster is at the top of a 72 m hill and weighs 966 N. The coaster (at this moment) has ____________ energy. Calculate it.

Gravitational Potential Energy
PE = m.g.h
note: weight = m. g = 966 N
so put the weight directly to the equation below
PE = 966 N . 72 m
PE = 69,552 N.m
PE = 69,552 Joule

7. What is the kinetic energy of a 3-kilogram ball that is rolling at 2 meters per second?

KE = ¹/₂ m v²
KE = ¹/₂ x 3 kg x (2 m/s)²
KE = ¹/₂ x 3 kg x 4 m²/s²
KE = 6 Joule

8. Object were lifted by a machine. One object had a mass of 2 kilograms if the Kinetic energy of the object was 1600 J. What was the velocity?

KE = ¹/₂ m v²
1,600 J = ¹/₂ x 2 kg x v²
v² = EK /(¹/₂ x 2 kg)
v² = 1,600 / 1
v = (1,600)^1/2
v = 40 m/s

9. Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.

KE = ¹/₂ m v²
KE = ¹/₂ x 1000 kg x (20 m/s)²
KE = ¹/₂ x 1000 kg x 400 m²/s²
KE = 200,000 Joule

INFINITE SERIES, POWER SERIES

1. The Geometric Series
     Example of geometric progressions;
a). 2, 4, 8, 16, 32, ...
b) 1, 2/3 , 4/9, 8/27, ...
c) a, ar, ar², ar³, ...

You may recall that in a geometric progression we multiply each term by some fixed number to get the next term.

The formula for the sum of n terms of the geometric progression is

(1.1)

That geometric series has a sum if and only if |r| < 1, and in this case the sum is

(1.2)

The series is then called convergent.

Problem, Section 1

1. In the bouncing ball example above, find the height of the tenth rebound, and the distance traveled by the ball after it touches the ground the tenth time. Compare this distance with the total distance traveled.(Suppose a bouncing ball rises each time to 2/3 of the height of the previous bounce then it would represent the heights of successive bounces in yards if the ball is originally dropped from a height of 1 yd).

 Answer:
a. The height of the tenth rebound
    1, 2/3 , 4/9, 8/27, ...
    U₁ = a = 1 (the originally height)
    U₂  = 2/3(the height of the first rebound)
    U₃  = 4/9(the height of the second rebound)
      '
      '
    U₁₁  = ? (the height of the eleventh rebound)
    U_n= a.r^n-1
    U₁₁= 1.(2/3)^11-1
    U₁₁= 1.(2/3)¹⁰
    U₁₁= 0.017 yd
    So, the height of the tenth rebound is 0.017 yd.

b. The distance traveled by the ball after it touches the ground the tenth time.

c. The total distance traveled.

2. Derive the formula (1.1) for the sum Sn of the geometric progression Sn = a+ar+ar² + · · · + ar^n-1. Hint: Multiply Sn by r and subtract the result from Sn; then solve for Sn. Show that the geometric series converges if and only if |r| < 1; also show that if |r| < 1, the sum is given by equation (1.2).

Use equation (1.2) to find the fractions that are equivalent to the following repeating decimals:
3. 0.55555 · · ·  
4. 0.818181 · · ·
5. 0.583333 · · ·
6. 0.61111 · · ·
7. 0.185185 · · ·
8. 0.694444 · · ·
9. 0.857142857142 · · ·
10. 0.576923076923076923 · · ·
11. 0.678571428571428571 · · ·