1. Travel Graphs

2. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t

3. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t
Distance = total amount travelled

4. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t
Distance = total amount travelled
Displacement = how far from the starting point

5. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t
Distance = total amount travelled
Displacement = how far from the starting point
(i) Find the height of the ball after 1 second

6. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t
Distance = total amount travelled
Displacement = how far from the starting point
(i) Find the height of the ball after 1 second
when t  1, x  5 1 8  1
 35

7. Travel Graphs
e.g. A ball is bounced and its distance from the ground is graphed.
x
80 x  5t  8  t 
60
40
20
2 4 6 8 t
Distance = total amount travelled
Displacement = how far from the starting point
(i) Find the height of the ball after 1 second
when t  1, x  5 1 8  1
 35
After 1 second the ball is 35 metres above the ground

8. (ii) At what other time is the ball this same height above the ground?

9. (ii) At what other time is the ball this same height above the ground?
when x  35,

10. (ii) At what other time is the ball this same height above the ground?
when x  35, 5t  8  t   35
t 8  t   7
8t  t 2  7
t 2  8t  7  0
 t  1 t  7   0
t  1 or t 7

11. (ii) At what other time is the ball this same height above the ground?
when x  35, 5t  8  t   35
t 8  t   7
8t  t 2  7
t 2  8t  7  0
 t  1 t  7   0
t  1 or t  7
 ball is 35 metres above ground again after 7 seconds

12. (ii) At what other time is the ball this same height above the ground?
when x  35, 5t  8  t   35
t 8  t   7
8t  t 2  7
t 2  8t  7  0
 t  1 t  7   0
t  1 or t  7
 ball is 35 metres above ground again after 7 seconds
change in displacement
Average velocity =
change in time

13. (ii) At what other time is the ball this same height above the ground?
when x  35, 5t  8  t   35
t 8  t   7
8t  t 2  7
t 2  8t  7  0
 t  1 t  7   0
t  1 or t  7
 ball is 35 metres above ground again after 7 seconds
change in displacement
Average velocity =
change in time
x2  x1

t2  t1

14. (iii) Find the average velocity during the 1st second

15. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35

16. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s

17. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s
(iv) Find the average velocity during the fifth second

18. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s
(iv) Find the average velocity during the fifth second
when t  4, x  5  4  8  4 
=80

19. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s
(iv) Find the average velocity during the fifth second
when t  4, x  5  4  8  4 
=80
when t  5, x  5  5  8  5 
=75

20. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s
(iv) Find the average velocity during the fifth second
x2  x1
when t  4, x  5  4  8  4  average velocity 
t2  t1
=80
75  80
when t  5, x  5  5  8  5  
54
=75  5

21. (iii) Find the average velocity during the 1st second
x2  x1
average velocity 
t2  t1
35  0

1 0
 35
 average velocity during the 1st second was 35m/s
(iv) Find the average velocity during the fifth second
x2  x1
when t  4, x  5  4  8  4  average velocity 
t2  t1
=80
75  80
when t  5, x  5  5  8  5  
54
=75  5
 average velocity during the 5th second was  5m/s

22. (iv) Find the average velocity during its 8 seconds in the air

23. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0

24. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s

25. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s
distance travelled
Average speed =
time taken

26. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s
distance travelled
Average speed =
time taken
(v) Find the average speed during its 8 seconds in the air

27. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s
distance travelled
Average speed =
time taken
(v) Find the average speed during its 8 seconds in the air
distance travelled
average speed 
time taken

28. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s
distance travelled
Average speed =
time taken
(v) Find the average speed during its 8 seconds in the air
distance travelled
average speed 
time taken
160

8
 20

29. (iv) Find the average velocity during its 8 seconds in the air
x2  x1
average velocity 
t2  t1
00

80
0
 average velocity during the 8 seconds was 0m/s
distance travelled
Average speed =
time taken
(v) Find the average speed during its 8 seconds in the air
distance travelled
average speed 
time taken
160

8
 20
 average speed during the 8 seconds was 20m/s

30. Applications of Calculus To
The Physical World

31. Applications of Calculus To
The Physical World
Displacement (x)
Distance from a point, with direction.

32. Applications of Calculus To
The Physical World
Displacement (x)
Distance from a point, with direction.
 v, dx , x 
Velocity  
 dt 
The rate of change of displacement with respect to time i.e. speed
with direction.

33. Applications of Calculus To
The Physical World
Displacement (x)
Distance from a point, with direction.
 v, dx , x 
Velocity  
 dt 
The rate of change of displacement with respect to time i.e. speed
with direction.
 dv d 2 x 
Acceleration  a, , 2 , , v 
x 
 dt dt 
The rate of change of velocity with respect to time

34. Applications of Calculus To
The Physical World
Displacement (x)
Distance from a point, with direction.
 v, dx , x 
Velocity  
 dt 
The rate of change of displacement with respect to time i.e. speed
with direction.
 dv d 2 x 
Acceleration  a, , 2 , , v 
x 
 dt dt 
The rate of change of velocity with respect to time
NOTE: “deceleration” or slowing down is when acceleration is in
the opposite direction to velocity.

35. Displacement
Velocity
Acceleration

36. Displacement
differentiate Velocity
Acceleration

37. Displacement
differentiate Velocity
Acceleration

38. Displacement
differentiate Velocity integrate
Acceleration

39. Displacement
differentiate Velocity integrate
Acceleration

40. x
2
1
1 2 3 4 t
-1
-2

41. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1
-2

42. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1
-2

x
1 2 3 4 t

43. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1
-2
slope=instantaneous acceleration

x
1 2 3 4 t

44. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1
-2
slope=instantaneous acceleration

x
1 2 3 4 t

x
1 2 3 4 t

45. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled
-2
slope=instantaneous acceleration

x
1 2 3 4 t

x
1 2 3 4 t

46. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration

x
1 2 3 4 t

x
1 2 3 4 t

47. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement

x
1 2 3 4 t

x
1 2 3 4 t

48. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement  1m

x
1 2 3 4 t

x
1 2 3 4 t

49. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement  1m

x
(iii) average speed
1 2 3 4 t

x
1 2 3 4 t

50. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement  1m

x
7
(iii) average speed  m/s
4
1 2 3 4 t

x
1 2 3 4 t

51. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement  1m

x
7
(iii) average speed  m/s
4
1 2 3 4 t (iv) average velocity

x
1 2 3 4 t

52. x
2 slope=instantaneous velocity
1
1 2 3 4 t
-1 e.g. (i) distance traveled  7 m
-2
slope=instantaneous acceleration (ii) total displacement  1m

x
7
(iii) average speed  m/s
4
1
1 2 3 4 t (iv) average velocity  m/s
4

x
1 2 3 4 t

53. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2

54. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.

55. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42

56. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42
b) Find the times when the particle is stationary.

57. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42
b) Find the times when the particle is stationary.
Particle is stationary when v = 0

58. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42
b) Find the times when the particle is stationary.
Particle is stationary when v = 0
i.e. 3t 2  42t  0

59. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42
b) Find the times when the particle is stationary.
Particle is stationary when v = 0
i.e. 3t 2  42t  0
3t t  14   0
t  0 or t  14

60. e.g. (i) The displacement x from the origin at time t seconds, of a
particle traveling in a straight line is given by the formula
x  t 3  21t 2
a) Find the acceleration of the particle at time t.
x  t 3  21t 2
v  3t 2  42t
a  6t  42
b) Find the times when the particle is stationary.
Particle is stationary when v = 0
i.e. 3t 2  42t  0
3t t  14   0
t  0 or t  14
Particle is stationary initially and again after 14 seconds

61. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.

62. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c

63. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0

64. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0
i.e. 0  0  0  c
c0

65. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0
i.e. 0  0  0  c
c0
 v  2t  3t 2

66. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0
i.e. 0  0  0  c
c0
 v  2t  3t 2
x  t2  t3  c

67. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0
i.e. 0  0  0  c
c0
 v  2t  3t 2
x  t2  t3  c
when t  0, x  7
i.e. 7  0  0  c
c7

68. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t
v  2t  3t 2  c
when t  0, v  0
i.e. 0  0  0  c
c0
 v  2t  3t 2
x  t2  t3  c
when t  0, x  7
i.e. 7  0  0  c
c7
 x  t2  t3  7

69. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t when t  3, x  32  33  7
v  2t  3t 2  c  43
when t  0, v  0
i.e. 0  0  0  c
c0
 v  2t  3t 2
x  t2  t3  c
when t  0, x  7
i.e. 7  0  0  c
c7
 x  t2  t3  7

70. (ii) A particle is moving on the x axis. It started from rest at t = 0 from
the point x = 7.
If its acceleration at time t is 2 + 6t find the position of the particle
when t = 3.
a  2  6t when t  3, x  32  33  7
v  2t  3t 2  c  43
when t  0, v  0
after 3 seconds the particle is 43
i.e. 0  0  0  c
units to the right of O.
c0
 v  2t  3t 2
x  t2  t3  c
when t  0, x  7
i.e. 7  0  0  c
c7
 x  t2  t3  7

71. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.

72. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?

73. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1

74. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin

75. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.

76. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
4 t 24
1 
t2 t2

77. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
4 t 24
1 
t2 t2
t2

t2

78. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
4 t 24
1 
t2 t2
t2
 4
t  2  x  1
t2

79. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
4 t 24 4  1
1  v
t2 t2 t  2
2
t2
 4
t  2  x  1
t2

80. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
4 t 24 4  1
1  v
t2 t2 t  2
2
t2
 4 v
4
t  2  x  1 t  2
2
t2

81. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
t 24 4  1 4  2  t  2  1
1
4
1  v a
t2 t2 t  2  t  2
2 4
t2
 4 v
4
t  2  x  1 t  2
2
t2

82. e.g. 2001 HSC Question 7c)
A particle moves in a straight line so that its displacement, in metres,
is given by t2
x
t2
where t is measured in seconds.
(i) What is the displacement when t = 0?
02
when t  0, x 
02
= 1
 the particle is 1 metre to the left of the origin
4
(ii) Show that x  1 
t2
Hence find expressions for the velocity and the acceleration in terms of t.
t 24 4  1 4  2  t  2  1
1
4
1  v a
t2 t2 t  2  t  2
2 4
t2 8
 4 v
4
a
t  2  x  1 t  2
2
 t  2
3
t2

83. (iii) Is the particle ever at rest? Give reasons for your answer.

84. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2

85. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest

86. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?

87. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?
lim v
t 

88. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?
4
lim v  lim
t 
 
t  t  2 2

89. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?
4
lim v  lim
t 
 
t  t  2 2
0

90. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?
4 v
lim v  lim OR
t 
 
t  t  2 2
4
v
0 1 t  2
2
t

91. (iii) Is the particle ever at rest? Give reasons for your answer.
4
v 2  0
t  2
 the particle is never at rest
(iv) What is the limiting velocity of the particle as t increases indefinitely?
4 v
lim v  lim OR
t 
 
t  t  2 2
4
v
0 1 t  2
2
t
 the limiting velocity of the particle is 0 m/s

92. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3

93. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2

94. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
amplitude  1 unit

95. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
amplitude  1 unit
shift  3 units

96. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2
amplitude  1 unit period 
2
shift  3 units 

97. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 

98. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 
x
4
3
2
1
  3  5 3 7 2 t
4 2 4 4 2 4

99. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 
x
4
3
2
1
  3  5 3 7 2 t
4 2 4 4 2 4

100. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 
x
4
3
2
1
  3  5 3 7 2 t
4 2 4 4 2 4

101. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 
x
4
3
2
1
  3  5 3 7 2 t
4 2 4 4 2 4

102. (ii) 2002 HSC Question 8b)
A particle moves in a straight line. At time t seconds, its distance x
metres from a fixed point O in the line is given by x  sin 2t  3
(i) Sketch the graph of x as a function of t for 0  t  2
2 
amplitude  1 unit period  divisions 
2 4
shift  3 units 
x
4 x  sin 2t  3
3
2
1
  3  5 3 7 2 t
4 2 4 4 2 4

103. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.

104. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.
Particle is at rest when velocity = 0

105. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.
Particle is at rest when velocity = 0
dx
 0 i.e. the stationary points
dt

106. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.
Particle is at rest when velocity = 0
dx
 0 i.e. the stationary points
dt

when t  seconds, x  4 metres
4
3
t seconds, x  2 metres
4
5
t seconds, x  4 metres
4
7
t seconds, x  2 metres
4

107. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.
Particle is at rest when velocity = 0
dx
 0 i.e. the stationary points
dt

when t  seconds, x  4 metres
4
3
t seconds, x  2 metres
4
5
t seconds, x  4 metres
4
7
t seconds, x  2 metres
4
(iii) Describe the motion completely.

108. (ii) Using your graph, or otherwise, find the times when the particle is at
rest, and the position of the particle at those times.
Particle is at rest when velocity = 0
dx
 0 i.e. the stationary points
dt

when t  seconds, x  4 metres
4
3
t seconds, x  2 metres
4
5
t seconds, x  4 metres
4
7
t seconds, x  2 metres
4
(iii) Describe the motion completely.
The particle oscillates between x=2 and x=4 with a period of
 seconds

109. Integrating Functions of Time

x
1 2 3 4 t

110. Integrating Functions of Time

x
4
change in displacement   xdt

0
1 2 3 4 t

111. Integrating Functions of Time

x
4
change in displacement   xdt

0
1 2 3 4 t 1 3 4
change in distance   xdt   xdt   xdt
  
0 1 3

112. Integrating Functions of Time

x
4
change in displacement   xdt

0
1 2 3 4 t 1 3 4
change in distance   xdt   xdt   xdt
  
0 1 3

x
1 2 3 4 t

113. Integrating Functions of Time

x
4
change in displacement   xdt

0
1 2 3 4 t 1 3 4
change in distance   xdt   xdt   xdt
  
0 1 3

x
4
change in velocity   dt
x
0
1 2 3 4 t

114. Integrating Functions of Time

x
4
change in displacement   xdt

0
1 2 3 4 t 1 3 4
change in distance   xdt   xdt   xdt
  
0 1 3

x
4
change in velocity   dt
x
0
1 2 3 4 t 2 4
change in speed    dt   dt
x x
0 2

115. Derivative Graphs
Function 1st derivative 2nd derivative
displacement velocity acceleration

116. Derivative Graphs
Function 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept

117. Derivative Graphs
Function 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept

118. Derivative Graphs
Function 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive

119. Derivative Graphs
Function 1st derivative 2nd derivative
displacement velocity acceleration
stationary point x intercept
inflection point stationary point x intercept
increasing positive
decreasing negative

120. Derivative Graphs
Function 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

121. Derivative Graphs
Function 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

122. graph type integrate differentiate

123. graph type integrate differentiate
horizontal line oblique line x axis

124. graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line

125. graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique line
inflects at turning pt

126. graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique line
inflects at turning pt
Remember:
• integration = area

127. graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique line
inflects at turning pt
Remember:
• integration = area
• on a velocity graph, total area = distance
total integral = displacement

128. graph type integrate differentiate
horizontal line oblique line x axis
oblique line parabola horizontal line
parabola cubic oblique line
inflects 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

129. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds

130. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?

131. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0

132. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0
2  4cos t  0
1
cos t 
2

133. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4
2  4cos t  0
1
cos t 
2

134. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4
2  4cos t  0 1
1 cos  
cos t  2
2 

3

135. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3

136. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3
 5
 particle is at rest after seconds and again after seconds
3 3

137. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3
 5
 particle is at rest after seconds and again after seconds
3 3
(ii) What is the maximum velocity of the particle during this period?

138. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3
 5
 particle is at rest after seconds and again after seconds
3 3
(ii) What is the maximum velocity of the particle during this period?
4  4 cos t  4

139. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3
 5
 particle is at rest after seconds and again after seconds
3 3
(ii) What is the maximum velocity of the particle during this period?
4  4 cos t  4
2  2  4 cos t  6

140. (ii) 2003 HSC Question 7b)
The velocity of a particle is given by v  2  4 cos t for 0  t  2 ,
where v is measured in metres per second and t is measured in seconds
(i) At what times during this period is the particle at rest?
v0 Q1, 4 t   , 2  
2  4cos t  0 1  5
1 cos   t ,
cos t  2 3 3
2 

3
 5
 particle is at rest after seconds and again after seconds
3 3
(ii) What is the maximum velocity of the particle during this period?
4  4 cos t  4
2  2  4 cos t  6
 maximum velocity is 6 m/s

141. (iii) Sketch the graph of v as a function of t for 0  t  2

142. (iii) Sketch the graph of v as a function of t for 0  t  2
amplitude  4 units

143. (iii) Sketch the graph of v as a function of t for
amplitude  4 units
shift  2 units
flip upside down

144. (iii) Sketch the graph of v as a function of t for 0  t  2
2
amplitude  4 units period 
shift  2 units 1
 2
flip upside down

145. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2

146. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2
v
6
5
4
3
2
1
-1
  3 2 t
-2
2 2

147. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2
v
6
5
4
3
2
1
-1
  3 2 t
-2
2 2

148. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2
v
6
5
4
3
2
1
-1
  3 2 t
-2
2 2

149. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2
v
6
5
4
3
2
1
-1
  3 2 t
-2
2 2

150. (iii) Sketch the graph of v as a function of t for 0  t  2
2 2
amplitude  4 units period  divisions 
shift  2 units 1 4
 2 
flip upside down 
2
v
6
5 v  2  4 cos t
4
3
2
1
-1
  3 2 t
-2
2 2

151. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

152. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

3 
distance =    2  4 cos t  dt    2  4 cos t  dt
0 
3

153. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

3 
distance =    2  4 cos t  dt    2  4 cos t  dt
0 
=  2t  4sin t    2t 3 4sin t 
0 
3 3

154. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

3 
distance =    2  4 cos t  dt    2  4 cos t  dt
0 
=  2t  4sin t    2t 3 4sin t 
0 
3 3
 2  4sin  
=  0  0    2  4sin    2  
 3 3

155. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

3 
distance =    2  4 cos t  dt    2  4 cos t  dt
0 
=  2t  4sin t    2t 3 4sin t 
0 
3 3
 2  4sin  
=  0  0    2  4sin    2  
 3 3
 2 4 3 
=2  2   
 3 2 

156. (iv) Calculate the total distance travelled by the particle between t = 0
and t = 

3 
distance =    2  4 cos t  dt    2  4 cos t  dt
0 
=  2t  4sin t    2t 3 4sin t 
0 
3 3
 2  4sin  
=  0  0    2  4sin    2  
 3 3
 2 4 3 
=2  2   
 3 2 
2
=4 3  metres
3

157. (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  t  5

158. (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  t  5
(i) Write down the time at which the velocity of the particle is a maximum

159. (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  t  5
(i) Write down the time at which the velocity of the particle is a maximum
v   adt

160. (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  t  5
(i) Write down the time at which the velocity of the particle is a maximum
v   adt
 adt is a maximum when t  2

161. (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  t  5
(i) Write down the time at which the velocity of the particle is a maximum
dv
v   adt OR v is a maximum when 0
dt
 adt is a maximum when t  2

162. (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  t  5
(i) Write down the time at which the velocity of the particle is a maximum
dv
v   adt OR v is a maximum when 0
dt
 adt is a maximum when t  2
 velocity is a maximum when t  2 seconds

163. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.

164. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”

165. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt

166. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt
But v   adt

167. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt
But v   adt
 We must solve  adt  0

168. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt
But v   adt
 We must solve  adt  0
i.e. when is area above the axis = area below

169. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt
But v   adt
 We must solve  adt  0
i.e. when is area above the axis = area below
By symmetry this would be at t = 4

170. (ii) At what time during the interval 0  t  5 is the particle furthest
from the origin? Give reasons for your answer.
Question is asking, “when is displacement a maximum?”
dx
x is a maximum when 0
dt
But v   adt
 We must solve  adt  0
i.e. when is area above the axis = area below
By symmetry this would be at t = 4
 particle is furthest from the origin at t  4 seconds

171. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6

172. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4

173. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4 h
distance   y0  4 yodd  2 yeven  yn 
3

174. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4 h
distance   y0  4 yodd  2 yeven  yn 
3
t 0 2 4
v 0 1 5

175. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4 h
distance   y0  4 yodd  2 yeven  yn 
3
1 1
t 0 2 4
v 0 1 5

176. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4 h
distance   y0  4 yodd  2 yeven  yn 
3
1 4 1
t 0 2 4
v 0 1 5

177. (iv) 2007 HSC Question 10a) dx
An object is moving on the x-axis. The graph shows the velocity, ,
dt
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 t  6
(i) Using Simpson’s rule, estimate the distance travelled between t = 0
and t = 4 h
distance   y0  4 yodd  2 yeven  yn 
3
1 4 1 2
 0  4 1  5
t 0 2 4 3
v 0 1 5
 6 metres

178. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?

179. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt

180. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds

181. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(iii) Estimate the time at which the object returns to the origin. Justify
your answer.

182. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(iii) Estimate the time at which the object returns to the origin. Justify
your answer.
Question is asking, “when is displacement = 0?”

183. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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

184. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0

185. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
i.e. when is area above the axis = area below

186. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area
from t = 5 to 6

187. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area
from t = 5 to 6
In part (i) we estimated area from t = 0 to 4 to be 6,

188. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area A4
from t = 5 to 6
In part (i) we estimated area from t = 0 to 4 to be 6,
 A4  6

189. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
a
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area A4 5
from t = 5 to 6
In part (i) we estimated area from t = 0 to 4 to be 6,
 A4  6
5a  6

190. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
a
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area A4 5
from t = 5 to 6
In part (i) we estimated area from t = 0 to 4 to be 6,
 A4  6 a  1.2
5a  6

191. (ii) The object is initially at the origin. During which time(s) is the
displacement decreasing?
dx
x is decreasing when 0
dt
 displacement is decreasing when t  5 seconds
(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  vdt  0
a
i.e. when is area above the axis = area below
By symmetry, area from t = 4 to 5 equals area A4 5
from t = 5 to 6
In part (i) we estimated area from t = 0 to 4 to be 6,
 A4  6 a  1.2
5a  6  particle returns to the origin when t  7.2 seconds

192. (iv) Sketch the displacement, x, as a function of time.

193. (iv) Sketch the displacement, x, as a function of time.
x
8.5
6
2 4 6 8 t

194. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
x
8.5
6
2 4 6 8 t

195. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
x
8.5
6
2 4 6 8 t

196. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
x
8.5
6
2 4 6 8 t

197. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5
 when t  5, x  8.5
x
8.5
6
2 4 6 8 t

198. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5
 when t  5, x  8.5
returns to x = 0 when t = 7.2
x
8.5
6
2 4 6 7.2 8 t

199. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5
 when t  5, x  8.5
returns to x = 0 when t = 7.2
x
v is steeper between t = 2 and 4
8.5
than between t = 0 and 2
6  particle covers more distance
between t  2 and 4
2 4 6 7.2 8 t

200. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5
 when t  5, x  8.5
returns to x = 0 when t = 7.2
x
v is steeper between t = 2 and 4
8.5
than between t = 0 and 2
6  particle covers more distance
between t  2 and 4
when t > 6, v is constant
t  when t  6, x is a straight line
2 4 6 7.2 8

201. (iv) Sketch the displacement, x, as a function of time.
object is initially at the origin
when t = 4, x = 6
by symmetry of areas t = 6, x = 6
Area of triangle = 2.5
 when t  5, x  8.5
returns to x = 0 when t = 7.2
x
v is steeper between t = 2 and 4
8.5
than between t = 0 and 2
6  particle covers more distance
between t  2 and 4
when t > 6, v is constant
t  when t  6, x is a straight line
2 4 6 7.2 8

202. (v) 2005 HSC Question 7b)
dx
The graph shows the velocity, dt , of a particle as a function of time.
Initially the particle is at the origin.

203. (v) 2005 HSC Question 7b)
dx
The graph shows the velocity, dt , of a particle as a function of time.
Initially the particle is at the origin.
(i) At what time is the displacement, x, from the origin a maximum?

204. (v) 2005 HSC Question 7b)
dx
The graph shows the velocity, dt , of a particle as a function of time.
Initially the particle is at the origin.
(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

205. (v) 2005 HSC Question 7b)
dx
The graph shows the velocity, dt , of a particle as a function of time.
Initially the particle is at the origin.
(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

206. (ii) At what time does the particle return to the origin? Justify
your answer

207. (ii) At what time does the particle return to the origin? Justify
your answer
Question is asking, “when is displacement = 0?”

208. (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?

209. (ii) At what time does the particle return to the origin? Justify
your answer
2 a w
a 2
Question is asking, “when is displacement = 0?”
i.e. when is area above the axis = area below?