12 x1 t04 06 integrating functions of time (2012)
TRANSCRIPT
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Integrating Functions of Time
t1 2 3 4
x
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t1 2 3 4
x
4
0
ntdisplacemein change dtx
Integrating Functions of Time
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t1 2 3 4
x
4
0
ntdisplacemein change dtx
4
3
3
1
1
0
distancein change dtxdtxdtx
Integrating Functions of Time
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t1 2 3 4
x
4
0
ntdisplacemein change dtx
4
3
3
1
1
0
distancein change dtxdtxdtx
t1 2 3 4
x
Integrating Functions of Time
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t1 2 3 4
x
4
0
ntdisplacemein change dtx
4
3
3
1
1
0
distancein change dtxdtxdtx
t1 2 3 4
x
4
0
yin velocit change dtx
Integrating Functions of Time
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t1 2 3 4
x
4
0
ntdisplacemein change dtx
4
3
3
1
1
0
distancein change dtxdtxdtx
t1 2 3 4
x
4
0
yin velocit change dtx
4
2
2
0
speedin change dtxdtx
Integrating Functions of Time
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive
decreasing negative
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive
decreasing negative
concave up increasing positive
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Derivative GraphsFunction 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive
decreasing negative
concave up increasing positive
concave down decreasing negative
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graph type integrate differentiate
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graph type integrate differentiate
horizontal line oblique line x axis
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graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
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graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique lineinflects at turning pt
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graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique lineinflects at turning pt
Remember:• integration = area
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graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique lineinflects at turning pt
Remember:• integration = area• on a velocity graph, total area = distance
total integral = displacement
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graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique lineinflects at turning pt
Remember:• integration = area• on a velocity graph, total area = distance
total integral = displacement• on an acceleration graph, total area = speed
total integral = velocity
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest?
0 2t
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest? 0v
0 2t
2 4cos 0t 1cos2
t
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest? 0v
0 2t
2 4cos 0t 1cos2
t
Q1, 4 1cos2
3
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest? 0v
0 2t
2 4cos 0t 1cos2
t
Q1, 4 1cos2
3
, 2t 5,
3 3t
5 particle is at rest after seconds and again after seconds3 3
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest? 0v
0 2t
2 4cos 0t 1cos2
t
Q1, 4 1cos2
3
, 2t 5,
3 3t
5 particle is at rest after seconds and again after seconds3 3
(ii) What is the maximum velocity of the particle during this period?
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(ii) 2003 HSC Question 7b)The velocity of a particle is given by for , where v is measured in metres per second and t is measured in seconds
2 4cosv t
(i) At what times during this period is the particle at rest? 0v
0 2t
2 4cos 0t 1cos2
t
Q1, 4 1cos2
3
, 2t 5,
3 3t
5 particle is at rest after seconds and again after seconds3 3
(ii) What is the maximum velocity of the particle during this period?
2 2 4cos 6t
maximum velocity is 6 m/s
4 4cos 4t
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(iii) Sketch the graph of v as a function of t for 0 2t
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
(iii) Sketch the graph of v as a function of t for 0 2t
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
(iii) Sketch the graph of v as a function of t for 0 2t
6v
-2
-1
4
2 3
2 2 t
1
23
5
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
(iii) Sketch the graph of v as a function of t for 0 2t
6v
-2
-1
4
2 3
2 2 t
1
23
5
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
(iii) Sketch the graph of v as a function of t for 0 2t
6v
-2
-1
4
2 3
2 2 t
1
23
5
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
(iii) Sketch the graph of v as a function of t for 0 2t
6v
-2
-1
4
2 3
2 2 t
1
23
5
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2period1
2
2divisions4
2
shift 2 unitsflip upside down
amplitude 4 units
2 4cosv t
(iii) Sketch the graph of v as a function of t for 0 2t
6v
-2
-1
4
2 3
2 2 t
1
23
5
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(iv) Calculate the total distance travelled by the particle between t = 0 and t =
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(iv) Calculate the total distance travelled by the particle between t = 0 and t =
3
03
distance = 2 4cos 2 4cost dt t dt
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(iv) Calculate the total distance travelled by the particle between t = 0 and t =
3
03
distance = 2 4cos 2 4cost dt t dt
0
3 3= 2 4sin 2 4sint t t t
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(iv) Calculate the total distance travelled by the particle between t = 0 and t =
3
03
distance = 2 4cos 2 4cost dt t dt
2=4 3 metres3
0
3 3= 2 4sin 2 4sint t t t
2= 0 0 2 4sin 2 4sin3 3
2 4 3=2 23 2
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(iii) 2004 HSC Question 9b)A particle moves along the x-axis. Initially it is at rest at the origin. The graph shows the acceleration, a, of the particle as a function of time t for 0 5t
(i) Write down the time at which the velocity of the particle is a maximum
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(iii) 2004 HSC Question 9b)A particle moves along the x-axis. Initially it is at rest at the origin. The graph shows the acceleration, a, of the particle as a function of time t for 0 5t
(i) Write down the time at which the velocity of the particle is a maximum
v adt is a maximum when 2adt t
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(iii) 2004 HSC Question 9b)A particle moves along the x-axis. Initially it is at rest at the origin. The graph shows the acceleration, a, of the particle as a function of time t for 0 5t
(i) Write down the time at which the velocity of the particle is a maximum
v adt is a maximum when 2adt t
OR is a maximum when 0dvvdt
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(iii) 2004 HSC Question 9b)A particle moves along the x-axis. Initially it is at rest at the origin. The graph shows the acceleration, a, of the particle as a function of time t for 0 5t
(i) Write down the time at which the velocity of the particle is a maximum
velocity is a maximum when 2 secondst
v adt is a maximum when 2adt t
OR is a maximum when 0dvvdt
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(ii) At what time during the interval is the particle furthest from the origin? Give reasons for your answer.
0 5t
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(ii) At what time during the interval is the particle furthest from the origin? Give reasons for your answer.
is a maximum when 0dxxdt
0 5t
Question is asking, “when is displacement a maximum?”
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(ii) At what time during the interval is the particle furthest from the origin? Give reasons for your answer.
is a maximum when 0dxxdt
0 5t
Question is asking, “when is displacement a maximum?”
But v adt We must solve 0adt
![Page 45: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/45.jpg)
(ii) At what time during the interval is the particle furthest from the origin? Give reasons for your answer.
is a maximum when 0dxxdt
0 5t
Question is asking, “when is displacement a maximum?”
But v adt We must solve 0adt
i.e. when is area above the axis = area belowBy symmetry this would be at t = 4
particle is furthest from the origin at 4 secondst
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(iv) 2007 HSC Question 10a)An object is moving on the x-axis. The graph shows the velocity, , of the object, as a function of t.The coordinates of the points shown on the graph are A(2,1), B(4,5), C(5,0) and D(6,–5). The velocity is constant for
dxdt
6t
(i) Using Simpson’s rule, estimate the distance travelled between t = 0 and t = 4
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(iv) 2007 HSC Question 10a)An object is moving on the x-axis. The graph shows the velocity, , of the object, as a function of t.The coordinates of the points shown on the graph are A(2,1), B(4,5), C(5,0) and D(6,–5). The velocity is constant for
dxdt
6t
(i) Using Simpson’s rule, estimate the distance travelled between t = 0 and t = 4 0distance 4 2
3 odd even nh y y y y
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(iv) 2007 HSC Question 10a)An object is moving on the x-axis. The graph shows the velocity, , of the object, as a function of t.The coordinates of the points shown on the graph are A(2,1), B(4,5), C(5,0) and D(6,–5). The velocity is constant for
dxdt
6t
(i) Using Simpson’s rule, estimate the distance travelled between t = 0 and t = 4
t 0 2 4v 0 1 5
0distance 4 23 odd even nh y y y y
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(iv) 2007 HSC Question 10a)An object is moving on the x-axis. The graph shows the velocity, , of the object, as a function of t.The coordinates of the points shown on the graph are A(2,1), B(4,5), C(5,0) and D(6,–5). The velocity is constant for
dxdt
6t
(i) Using Simpson’s rule, estimate the distance travelled between t = 0 and t = 4
t 0 2 4v 0 1 5
1 14 0distance 4 2
3 odd even nh y y y y
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(iv) 2007 HSC Question 10a)An object is moving on the x-axis. The graph shows the velocity, , of the object, as a function of t.The coordinates of the points shown on the graph are A(2,1), B(4,5), C(5,0) and D(6,–5). The velocity is constant for
dxdt
6t
(i) Using Simpson’s rule, estimate the distance travelled between t = 0 and t = 4
t 0 2 4v 0 1 5
1 14 2 0 4 1 536 metres
0distance 4 23 odd even nh y y y y
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(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
![Page 52: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/52.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
![Page 53: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/53.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.
![Page 54: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/54.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
![Page 55: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/55.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
![Page 56: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/56.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area below
![Page 57: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/57.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area belowBy symmetry, area from t = 4 to 5 equals areafrom t = 5 to 6
![Page 58: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/58.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area belowBy symmetry, area from t = 4 to 5 equals areafrom t = 5 to 6In part (i) we estimated area from t = 0 to 4 to be 6,
![Page 59: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/59.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area belowBy symmetry, area from t = 4 to 5 equals areafrom t = 5 to 6In part (i) we estimated area from t = 0 to 4 to be 6,
4 6A
4A
![Page 60: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/60.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area belowBy symmetry, area from t = 4 to 5 equals areafrom t = 5 to 6In part (i) we estimated area from t = 0 to 4 to be 6,
4 6A
4A
a
5
5 6a 1.2a
![Page 61: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/61.jpg)
(ii) The object is initially at the origin. During which time(s) is the displacement decreasing?
is decreasing when 0dxxdt
displacement is decreasing when 5 secondst
(iii) Estimate the time at which the object returns to the origin. Justify your answer.Question is asking, “when is displacement = 0?”
But x vdt We must solve 0vdt
i.e. when is area above the axis = area belowBy symmetry, area from t = 4 to 5 equals areafrom t = 5 to 6In part (i) we estimated area from t = 0 to 4 to be 6,
4 6A
4A
a
5
5 6a 1.2a particle returns to the origin when 7.2 secondst
![Page 62: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/62.jpg)
(iv) Sketch the displacement, x, as a function of time.
![Page 63: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/63.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
![Page 64: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/64.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the origin
![Page 65: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/65.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
![Page 66: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/66.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
![Page 67: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/67.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5when 5, 8.5t x
![Page 68: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/68.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5when 5, 8.5t x
returns to x = 0 when t = 7.2
7.2
![Page 69: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/69.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5when 5, 8.5t x
returns to x = 0 when t = 7.2
7.2
v is steeper between t = 2 and 4 than between t = 0 and 2
particle covers more distance between 2 and 4t
![Page 70: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/70.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5when 5, 8.5t x
returns to x = 0 when t = 7.2
7.2
v is steeper between t = 2 and 4 than between t = 0 and 2
particle covers more distance between 2 and 4t
when t > 6, v is constantwhen 6, is a straight linet x
![Page 71: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/71.jpg)
(iv) Sketch the displacement, x, as a function of time.
x
t2 4 6 8
6
8.5
object is initially at the originwhen t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5when 5, 8.5t x
returns to x = 0 when t = 7.2
7.2
v is steeper between t = 2 and 4 than between t = 0 and 2
particle covers more distance between 2 and 4t
when t > 6, v is constantwhen 6, is a straight linet x
![Page 72: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/72.jpg)
(v) 2005 HSC Question 7b)
The graph shows the velocity, , of a particle as a function of time.Initially the particle is at the origin.
dxdt
(i) At what time is the displacement, x, from the origin a maximum?
![Page 73: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/73.jpg)
(v) 2005 HSC Question 7b)
The graph shows the velocity, , of a particle as a function of time.Initially the particle is at the origin.
dxdt
(i) At what time is the displacement, x, from the origin a maximum?
Displacement is a maximum when area is most positive, also when velocity is zero
i.e. when t = 2
![Page 74: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/74.jpg)
(ii) At what time does the particle return to the origin? Justify your answer
![Page 75: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/75.jpg)
(ii) At what time does the particle return to the origin? Justify your answer
Question is asking, “when is displacement = 0?”
i.e. when is area above the axis = area below?
![Page 76: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/76.jpg)
(ii) At what time does the particle return to the origin? Justify your answer
Question is asking, “when is displacement = 0?”
i.e. when is area above the axis = area below?
2 a
a 2
w
![Page 77: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/77.jpg)
(ii) At what time does the particle return to the origin? Justify your answer
Question is asking, “when is displacement = 0?”
i.e. when is area above the axis = area below?
2 a
a 2
w
2w = 2w = 1
![Page 78: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/78.jpg)
(ii) At what time does the particle return to the origin? Justify your answer
Question is asking, “when is displacement = 0?”
i.e. when is area above the axis = area below?
2 a
a 2
w
2w = 2w = 1
Returns to the origin after 4 seconds
![Page 79: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/79.jpg)
2
2(iii) Draw a sketch of the acceleration, , as afunction of
time for 0 6
d xdt
t
t1 2 3 5 6
2
2d xdt
![Page 80: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/80.jpg)
2
2(iii) Draw a sketch of the acceleration, , as afunction of
time for 0 6
d xdt
t
t1 2 3 5 6
2
2d xdt
you get the xaxis
differentiate a horizontal line
![Page 81: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/81.jpg)
2
2(iii) Draw a sketch of the acceleration, , as afunction of
time for 0 6
d xdt
t
t1 2 3 5 6
2
2d xdt
you get the xaxis
differentiate a horizontal line
from 1 to 3 we have a cubic, inflects at 2, and is decreasing
differentiate, you get a parabola, stationary at 2, it is below the x axis
![Page 82: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/82.jpg)
2
2(iii) Draw a sketch of the acceleration, , as afunction of
time for 0 6
d xdt
t
t1 2 3 5 6
2
2d xdt
you get the xaxis
differentiate a horizontal line
from 1 to 3 we have a cubic, inflects at 2, and is decreasing
differentiate, you get a parabola, stationary at 2, it is below the x axis
from 5 to 6 is a cubic, inflects at 6 and is increasing (using symmetry)
differentiate, you get a parabola stationary at 6, it is above the x axis
![Page 83: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/83.jpg)
2
2(iii) Draw a sketch of the acceleration, , as afunction of
time for 0 6
d xdt
t
t1 2 3 5 6
2
2d xdt
you get the xaxis
differentiate a horizontal line
from 1 to 3 we have a cubic, inflects at 2, and is decreasing
differentiate, you get a parabola, stationary at 2, it is below the x axis
from 5 to 6 is a cubic, inflects at 6 and is increasing (using symmetry)
differentiate, you get a parabola stationary at 6, it is above the x axis
![Page 84: 12 x1 t04 06 integrating functions of time (2012)](https://reader033.vdocuments.us/reader033/viewer/2022052523/556f5e70d8b42a916b8b5112/html5/thumbnails/84.jpg)
Exercise 3C; 1 ace etc, 2 ace etc, 4a, 7ab(i), 8, 9a, 10, 13, 16, 18