2.
The first thing we must do to solve this
problem is find how long Wilfred and John will
be on the ride before Francis and Kieran get
off.
3.
The first thing we must do to solve this
problem is find how long Wilfred and John will
be on the ride before Francis and Kieran get
off.
We can do this by subtracting the time of
Wilfred and John’s boarding of the ride from
the time when Francis and Kieran get off.
4.
The first thing we must do to solve this
problem is find how long Wilfred and John will
be on the ride before Francis and Kieran get
off.
We can do this by subtracting the time of
Wilfred and John’s boarding of the ride from
the time when Francis and Kieran get off.
35
Time that Francis and
Kieran will get off
5.
The first thing we must do to solve this
problem is find how long Wilfred and John will
be on the ride before Francis and Kieran get
off.
We can do this by subtracting the time of
Wilfred and John’s boarding of the ride from
the time when Francis and Kieran get off.
35 - 21
Time that Francis and How long ‘F’ and ‘K’
Kieran will get off have been on the ride
when ‘W’ and ‘J’ get on
6.
The first thing we must do to solve this
problem is find how long Wilfred and John will
be on the ride before Francis and Kieran get
off.
We can do this by subtracting the time of
Wilfred and John’s boarding of the ride from
the time when Francis and Kieran get off.
35 - 21 = 14
Time that Francis and How long ‘F’ and ‘K’
Kieran will get off have been on the ride
when ‘W’ and ‘J’ get on
7.
This tells us that Francis and Kieran
will be getting off the ferris wheel
in 14 minutes.
8.
This tells us that Francis and Kieran
will be getting off the ferris wheel
in 14 minutes.
Therefore we can conclude that
Wilfred and John will have been on
the ferris wheel for 14 minutes.
9.
Now that we know how long they have been on
the ride, we must find what height Wilfred
and John are at when Francis and Kieran get
off.
10.
Now that we know how long they have been on
the ride, we must find what height Wilfred
and John are at when Francis and Kieran get
off.
We can do this by solving one of the equations
that we derived a few minutes ago with t=14
substituted into it.
11.
Now that we know how long they have been on
the ride, we must find what height Wilfred
and John are at when Francis and Kieran get
off.
We can do this by solving one of the equations
that we derived a few minutes ago with t=14
substituted into it.
***I will be using both the sine and
cosine equation for this part.***
30.
This means that Wilfred and John
were approximately 117 m off the
ground when Francis and Kieran got
off.
31.
We have just answered the first part of the
question.
32.
We have just answered the first part of the
question.
Now we must use this to answer the second
part of the question:
33.
We have just answered the first part of the
question.
Now we must use this to answer the second
part of the question:
How much longer will it be until
they reach the same height again?
34.
Some people might be confused by this
question, so lets take a look at it.
35.
Some people might be confused by this
question, so lets take a look at it.
This is the graph that we constructed earlier.
36.
Some people might be confused by this
question, so lets take a look at it.
The point on the graph is the point that we
just found. (t=14)
37.
Some people might be confused by this
question, so lets take a look at it.
The height of this point is 117m.
38.
Some people might be confused by this
question, so lets take a look at it.
However, there is another point with a height
of 117m.
39.
Some people might be confused by this
question, so lets take a look at it.
Therefore there is another time that they
will reach a height of 117m.
40.
So, we are looking to find that second time
which has a height of 117.3173m.
We can do this by making the equation equal
to 117.3173m and solving it.
41.
So, we are looking to find that second time
which has a height of 117.3173m.
We can do this by making the equation equal
to 117.3173m and solving it.
***This time I will only use the Sine
equation and explain the difference
between the two methods
afterwards.***
43.
Sine Equation
The Original Equation:
H(t)=75sin((2π/35)(t-10.75))+76
44.
Sine Equation
Substitute H(t)=117.3173
into the equation.
117.3173=75sin((2π/35)(t-10.75))+76
45.
Sine Equation
Let ((2π/35)(t-10.7))=Ø
to make it a little easier
to work with.
117.3173=75sinØ+76
46.
Sine Equation
Simplify the equation.
41.3173=75sinØ
47.
Sine Equation
Simplify the equation.
0.5509=sinØ
48.
Sine Equation
Calculate the arc sine
of 0.5509 [to four
decimal places].
0.5834=Ø
49.
Sine Equation
Now we have an angle.
The measure of this
angle is 0.5834 radians
Another thing we know
about this angle is that
it’s sine is positive.
WHY IS THIS
HELPFUL???
50.
Sine Equation
Well, when the sine of
an angle is positive, it
must either be in
Quadrant 1 or 2.
This particular angle is
in Quadrant 1.
HOW DO WE KNOW
THIS???
51.
Sine Equation
Here we have the Unit Circle.
QII QI
!/2
0
!
2!
3!/2
QIII QIV
52.
Sine Equation
The angle that separates
Quadrants I and II is π/2.
[or 1.5707...]
QII QI
!/2
0
!
2!
3!/2
QIII QIV
53.
Sine Equation
0.5834 is much less than
1.5708 so it lies in Quadrant I.
QII QI
!/2
0
!
2!
3!/2
QIII QIV
54.
Sine Equation
This means that to find the
other point with the exact
same sine value, it’s angle
would have to be in Quadrant
II.
55.
Sine Equation
This means that to find the
other point with the exact
same sine value, it’s angle
would have to be in Quadrant
II.
To find this angle we simply
subtract our angle from π (End
of Quadrant II).
56.
Sine Equation
This means that to find the
other point with the exact
same sine value, it’s angle
would have to be in Quadrant
II.
To find this angle we simply
subtract our angle from π (End
of Quadrant II).
π - Ø = Ø’
57.
Sine Equation
Here we have our
Quadrant II angle.
2.5582=Ø’
58.
Sine Equation
Now the two angles we
have are related
angles. This means
their sine values are
equivalent.
sinØ=sinØ’
59.
Sine Equation
And if we take the arc
sine of both sides and
we see that:
Ø=Ø’
60.
Sine Equation
Now we can continue
where we left off.
Since...
Ø=2.5582
66.
Cosine Equation
The Cosine equation works quite
similarly to that of Sine.
67.
Cosine Equation
The Cosine equation works quite
similarly to that of Sine.
The equation is a little bit
different as we saw earlier but
that’s about all that applies to
the first part of the procedure.
68.
Cosine Equation
One of the major differences is
when you simplify down to cosØ.
69.
Cosine Equation
One of the major differences is
when you simplify down to cosØ.
First of all, this value is negative
which means the angle is either in
Quadrant II or III.
70.
Cosine Equation
One of the major differences is
when you simplify down to cosØ.
First of all, this value is negative
which means the angle is either in
Quadrant II or III.
This one is in Quadrant II
because it is 2.1542 which is less
than π (3.1415), the end of
Quadrant II.
71.
Cosine Equation
This means Ø’ is in Quadrant III
and we must subtract Ø (angle
that passes Q I and into II) from
the full circle (2π) so it passes
through Q IV and into III.
72.
Cosine Equation
This means Ø’ is in Quadrant III
and we must subtract Ø (angle
that passes Q I and into II) from
the full circle (2π) so it passes
through Q IV and into III.
After that, continue as shown in
the Sine procedure until you get
t=25 as your answer.
73.
From here, we now can finish the problem off.
t=25 minutes
74.
Since it takes 14 minutes to get from the
start to 117.3173 m the first time.
14
75.
Since it takes 14 minutes to get from the
start to 117.3173 m the first time...
...And it takes 25 minutes to get from the start
to 117.3173 m the second time...
14 - 25
76.
...Then the time it takes to get from 117.3173
the first time to 117.3173 m the second time
must equal...
14 - 25 =?
78.
When Francis and Kieran get off
“The Wheel”, Wilfred and John will
be 117.3173 m from the ground.
It will take them 11 more minutes
to get to that exact same height
again.
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