1.12 a) At t = 0, the object is at x = A. v = B + 2Ct + 3Dt 2.
a = 2C + 6Dt .
b) At t = 0, the object is at x = A .
v = - ωA sin(ω t) .
a = - ω
2A cos(ω t).
1.13 a)
v = c t + v0 .
x = l
2 c t 2 + v0t + x0 .
b) v = -
l
2 kt 2 + v0 .
x = - l
6 kt 3 + v0t + x0 .

v
2
R ,
so if v is doubled, F will be four times as great.
String breaks Object falls to ground
This is a top view. In fact the object
falls to the ground.
b) The molecular forces that hold the wheel together.
c) The gravitational force exerted on the satellite by the earth.
2.5 a)
F = 20.00 N
N
W
Horizontal component of total force = F to the right
Horizontal acceleration = F
m
= 10.00 m. s
-2 .
b) Vertical component of total force = N - W upwards
The vertical acceleration is zero so the vertical component of the total force must be zero,
i. e. N - W = 0 or N = W.
c)
F = 20.00 N
N
W Friction = 10.0 N
Horizontal component of total force = 20.00 N - 10.00 N
= 10.00 N.
Horizontal acceleration = 5.00 m. s
-2.

c)
W
W cosθ
W sinθ
The downhill component is W sinθ .
2.7 a)
F = 20.00 N
N
W
θ = 30°
Horizontal component of total force,
FH = F cos(θ )
= 20.00 N cos(30°)
= 17.32 N .
Vertical component of total force, FV = N + F cos (90°- θ ) - W .
b) Horizontal component of acceleration =
FH
m = 8.66 m. s
-2.
The vertical acceleration is zero because the vertical component of the pulling force,
20.00 N cos (90° - 30°) = 10.00 N, is smaller than the weight, 2.00 × 9.8 N = 19.6 N. The contact
force N adjusts as FV changes and is just sufficient to prevent a downward acceleration.
( ... continued over)
Answers 110
c) If F = 40.00 N in the same direction, the vertical component of the pulling force would be 20.00 N
which is greater than the weight, 19.6 N. The upward acceleration would be 0.2 m. s
-2. The contact
force disappears.
d) Only the horizontal motion would be changed. The horizontal acceleration would be less.
2.8 a) Total force component ↑ = 4.00 N + 8.00 N cos (90° + 30°) = 0 N.
Total force component → = 8.00 N cos 30° = 6.93 N.
b) A single force of 6.93 N in the ← direction or any number of forces which combine to give a single
force equal to this.
2.9 a)
T
θ
θ
T
P
The magnitude of the force on the car equals the tension in the rope, T (and so is the magnitude of
the force on the tree). If the car has not quite started to move, the resultant force on the piece of rope at
the bend in the middle is zero. Taking force components perpendicular to the applied force:
P - T cos(90°-θ ) - T cos(90°-θ ) = 0 .
This gives T = P
2cos(90°-θ )
.
.
3.4 a)
N
x
300 N
500 N
200 N
1.8 m

3.8 a) About 0.75 m. b) About 10 m.
3.9 Buoyant force = weight of displaced air
= density of air × volume × g
= ρa
Vg .
Weight of balloon = density of helium × volume × g
= ρh
Vg .
Downward force required = ρa
Vg - ρ h
Vg = (ρa - ρh)Vg
= 10.8 N.
3.10

-l.
4.2 i) The other vertical forces are the buoyant force and the combined weight of the person and parachute.
ii)
Weight Drag force
Buoyant force
At terminal velocity, the total downward force is zero.
so weight - buoyant force - λv
T
2 = 0
vT = weight - buoyant force
λ
Answers 114
(... continued over)
iii) In the sky diving position: vT = 62 m. s-l ( = 220 km. h-l).
With parachute opened: vT = 5.7 m. s-l (= 20 km. h-l).
iv) Average downward acceleration = 5.7 m. s
- l - 62 m. s
-l
l s
≈ - 0.06 × 103 m. s
-2.

detailed shape is unimportant as long as the acceleration component decreases.)
Answers 115
Downward acceleration
Time
g
The downward velocity component as a function of time looks like this.
A
B
C
D
E
Time
Downward velocity
Since the downward acceleration is continually decreasing, the slope of the velocity curve decreases
with time.
The area ABC in the graph represents the height reached by the object. The area CDE will, of
necessity, equal the area ABC. (Why?)
graph, the time
taken to fall (CE) is greater than the time taken to rise (AC).
4.5 i) As t → ∞, e
-(λ/m )t → 0 so v → mg
λ .
ii) At t = 0 , e
-(λ/m )t = 1 and v = 0 .
iii) Now mg
λ - v = mg
λ e
-(λ/m )t
= e
-l 
  mg 
λ
when t = m
λ .
iv)
mg
λ
v
t
mg
λ e
-1×
m
λ
Answers 116
v) Since d
d t
e
-(λ/m )t = - λ
m e
-(λ/m )t
then a = mg
λ λ
m e
-(λ/m )t = g e
-(λ/m )t .
vi)
a
g
e
m
λ t
vii) mg - λv = mg e
-(λ/m )t
= ma = m
dv
dt .
viii) Terminal velocity occurs when
mg - λv
T = 0 ;
i. e when vT = mg
λ as in (i).
ix) The difference, v
T - v = mg
λ e
-(λ/m )t
x) d a
d t = - λ
m g e
-(λ/m )t = - λ
m a .
∴ m
d a
d t = - λ a .

When you consider that just 1 terabyte can hold 337,920 copies of War and Peace - one of the longest novels ever written - you can start to get a sense of just how insane that is. The previous record-holder was reportedly a 13-gigabyte proof, published in 2014.

So what’s this ridiculous maths problem? It’s been named the Boolean Pythagorean Triples problem, and was first posed by California-based mathematician Ronald Graham back in the 1980s.  

The problem centres around the Pythagorean formula a2 + b2 = c2, where a and b are the shorter sides of a triangle, and c is the hypotenuse, or longest side.  

Certain sets of three positive integers known as Pythagorean triples can be inserted into the formula, such as 32 + 42 = 52, 52 + 122 = 132, and 82 + 152 = 172.  

With this in mind, imagine that every integer is painted either red or blue.  

Graham asked if it's possible to colour all the integers either red or blue, so that no set of Pythagorean triples - a, b and c - were all the same colour. He put $100 on the line for anyone who could solve the problem. (That should cover a 1 terabyte drive.)

Andrew Moseman over at Popular Mechanics explains why that $100 looks mighty meagre, given the task ahead:

"What makes it so hard is that one integer can be part of multiple Pythagorean triples. Take 5. So 3, 4, and 5 are a Pythagorean triple. But so are 5, 12, and 13. If 5 is blue in the first example, then it has to be blue in the second, meaning either 12 or 13 has to be red.  

Carry this logic forward into much bigger numbers and you could see where this would start to get tricky. If 12 has to be red in that 5-12-13 triple, it might force changes down the line that would result in a monochrome triple somewhere."

Mathematicians Marijn Heule from the University of Texas, Victor Marek from the University of Kentucky, and Oliver Kullmann from Swansea University in the UK teamed up to figure it out, feeding a number of different techniques into University of Texas’s Stampede supercomputer and letting it narrow down the number of colour combination possibilities from 102,300 trillion (that’s 102,300) to just 1 trillion.

The 800-processor-strong supercomputer then took two days to mull over the remaining trillion and come up with a solution: 7,824. As soon as you try 7,825 or more integers, you can’t create the pattern Graham was looking for.

Guess who’s now $100 richer… split three ways! Good guy Graham came through with the cheque earlier this month.

The proof, which in maths means a written deductive argument that shows how you came to your answer, takes up a 200-terabyte file on the supercomputer - roughly equivalent to all the digitised text held by the US Library of Congress.  

According to Evelyn Lamb at Nature, the trio has created a 68-gigabyte compressed version of their solution, which would take about 30,000 hours to download, reconstruct and verify. The problem? No human could hope to read such a thing.  

Instead, the team had to employ another computer program to verify the results, and show that their solution meets the criteria of the original question, and Graham was satisfied with the confirmation.  

But critics are asking if this is enough. If no human can read the solution, that doesn’t mean it’s not correct, but it doesn’t address one very important component of mathematical problem-solving - it can’t explain why the colouring is impossible from 7,825 upwards, it just knows it is.

"Although the computer solution has cracked the Boolean Pythagorean triples problem, it hasn’t provided an underlying reason why the colouring is impossible, or explored whether the number 7,825 is meaningful," Lamb explains. "That echoes a common philosophical objection to the value of computer-assisted proofs: they may be correct, but are they really mathematics?"

If mathematics is about advancing human knowledge and understanding what numbers mean to us and the Universe around us, a computer churning out solutions that we’ll never understand seems to go against the very principles of the discipline.

As you ponder that question, you can take a look at the paper on pre-print website, arXiv.org. It hasn’t been peer-reviewed yet, because, well, we’d need a team of maths robots for that, I guess.

Learn More

Some Applications of the Wreath Product Construction

Charles Wells, The American Mathematical Monthly, 2018

Virtual currency "Riecoin" uses computations to help solve math problem

by Bob Yirka et al., TechXplore.com

Pi has been calculated out to 31.4 trillion decimals, Google announces on Pi Day

by Brett Molina et al., TechXplore.com

A traffic control algorithm for Clos network type ATM switches

S-H Choi et al., Journal of the Operational Research Society, 2017

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لا تفوت الفرصة ، استثمر في أسهم أرامكو. فقط 200 دولار كحد أدنى للإ…

المستثمر الذكي

Explanation:

answer; belief perseverance;

answer; in stage 6; morality of individual principles and conscience;