1. Velocity & Acceleration in
Terms of x

2. Velocity & Acceleration in
Terms of x
If v = f(x);

3. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 

4. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 
Proof: d 2 x dv
2

dt dt

5. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 
Proof: d 2 x dv
2

dt dt
dv dx
 
dx dt

6. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 
Proof: d 2 x dv
2

dt dt
dv dx
 
dx dt
dv
 v
dx

7. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 
Proof: d 2 x dv
2

dt dt
dv dx
 
dx dt
dv
 v
dx
dv d  1 2 
   v 
dx dv  2 

8. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
  v 
dt 2
dx  2 
Proof: d 2 x dv
2

dt dt
dv dx
 
dx dt
dv
 v
dx
dv d  1 2 
   v 
dx dv  2 
  v2 
d 1
 
dx  2 

9. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.

10. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 

11. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
1 2
v  3x  x 2  c
2

12. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
1 2
v  3x  x 2  c
2
when x  1, v  2
1 2
i.e. 2   31  12  c
2
c0

13. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
1 2
v  3x  x 2  c
2
when x  1, v  2
1 2
i.e. 2   31  12  c
2
c0
v2  6x  2x2

14. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
1 2
v  3x  x 2  c
2
when x  1, v  2
1 2
i.e. 2   31  12  c
2
c0
v2  6x  2x2
v   6x  2x2

15. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
NOTE:
1 2
v  3x  x 2  c
2 v2  0
when x  1, v  2
1 2
i.e. 2   31  12  c
2
c0
v2  6x  2x2
v   6x  2x2

16. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
NOTE:
1 2
v  3x  x 2  c
2 v2  0
when x  1, v  2 6x  2x2  0
1 2
i.e. 2   31  12  c
2
c0
v2  6x  2x2
v   6x  2x2

17. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
NOTE:
1 2
v  3x  x 2  c
2 v2  0
when x  1, v  2 6x  2x2  0
1 2
i.e. 2   31  12  c 2 x3  x   0
2
c0
v2  6x  2x2
v   6x  2x2

18. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
NOTE:
1 2
v  3x  x 2  c
2 v2  0
when x  1, v  2 6x  2x2  0
1 2
i.e. 2   31  12  c 2 x3  x   0
2
0 x3
c0
v2  6x  2x2
v   6x  2x2

19. e.g. (i) A particle moves in a straight line so that   3  2 x
x
Find its velocity in terms of x given that v = 2 when x = 1.
d 1 2 
 v   3  2x
dx  2 
NOTE:
1 2
v  3x  x 2  c
2 v2  0
when x  1, v  2 6x  2x2  0
1 2
i.e. 2   31  12  c 2 x3  x   0
2
0 x3
c0
Particle moves between x = 0
v  6x  2x
2 2
and x = 3 and nowhere else.
v   6x  2x2

20. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.

21. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2 
 v   3x 2
dx  2 

22. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2 
 v   3x 2
dx  2 
1 2
v  x3  c
2

23. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2 
 v   3x 2
dx  2 
1 2
v  x3  c
2
when t  0, x  1, v   2
i.e.
1
 2 2  13  c
2
c0

24. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2 
 v   3x 2
dx  2 
1 2
v  x3  c
2
when t  0, x  1, v   2
i.e.
1
 2 2  13  c
2
c0
 v 2  2 x3
v   2 x3

25. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2 
 v   3x 2 dx
  2x 3
dx  2  dt
1 2
v  x3  c
2
when t  0, x  1, v   2
i.e.  2   13  c
1 2
2
c0
 v 2  2 x3
v   2 x3

26. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2  (Choose –ve to satisfy
 v   3x 2 dx
  2x the initial conditions)
3
dx  2  dt
1 2
v  x3  c
2
when t  0, x  1, v   2
i.e.  2   13  c
1 2
2
c0
 v 2  2 x3
v   2 x3

27. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2  (Choose –ve to satisfy
 v   3x 2 dx
  2x the initial conditions)
3
dx  2  dt
1 2
v  x3  c
3
2   2x 2
when t  0, x  1, v   2
i.e.  2   13  c
1 2
2
c0
 v 2  2 x3
v   2 x3

28. (ii) A particle’s acceleration is given by   3x 2 Initially, the particle is
x .
1 unit to the right of O, and is traveling with a velocity of 2m/s in
the negative direction. Find x in terms of t.
d 1 2  (Choose –ve to satisfy
 v   3x 2 dx
  2x the initial conditions)
3
dx  2  dt
1 2
v  x3  c
3
2   2x 2
when t  0, x  1, v   2 3
dt 1 2

i.e.
1
 2 2  13  c dx 2
x
2
c0
 v 2  2 x3
v   2 x3

29. 3
dt 1 2
 x
dx 2

30. 3
dt 1 2
 x
dx 2
1
1 
t  2 x 2  c
2
1

 2x 2
c
2
 c
x

31. 3
dt 1 2
 x
dx 2
1
1 
t  2 x 2  c
2
1

 2x 2
c
2
 c
x
when t = 0, x = 1

32. 3
dt 1 2
 x
dx 2
1
1 
t  2 x 2  c
2
1

 2x 2
c
2
 c
x
when t = 0, x = 1
i.e. 0  2  c
c 2

33. 3
dt 1 2
 x
dx 2
1
1 
t  2 x 2  c
2
1

 2x 2
c
2
 c
x
when t = 0, x = 1
i.e. 0  2  c
c 2
2
t  2
x

34. 3
dt 1 2
 x
dx 2
1
1 
t  2 x 2  c OR
2
1

 2x 2
c
2
 c
x
when t = 0, x = 1
i.e. 0  2  c
c 2
2
t  2
x

35. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1

 2x 2
c
2
 c
x
when t = 0, x = 1
i.e. 0  2  c
c 2
2
t  2
x

36. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1 x

 2x 2
c 1   
1
  2 x 
2
2 2 1
 c
x
when t = 0, x = 1
i.e. 0  2  c
c 2
2
t  2
x

37. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1 x

 2x 2
c 1   
1
  2 x 
2
2 2 1
 c
x  1  1
when t = 0, x = 1  2 
 x 
i.e. 0  2  c
c 2
2
t  2
x

38. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1 x

 2x 2
c 1   
1
  2 x 
2
2 2 1
 c
x  1  1
when t = 0, x = 1  2 
 x 
2
i.e. 0  2  c t 2
x
c 2
2
t  2
x

39. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1 x

 2x 2
c 1   
1
  2 x 
2
2 2 1
 c
x  1  1
when t = 0, x = 1  2 
 x 
2
i.e. 0  2  c t 2
x
c 2
 t  2 
2 2
2
t  2 x
x

40. 3
dt 1 2
 x
dx 2
1 x 3
1  1 
2
t  2 x 2  c OR t x 2 dx
2 1
1 x

 2x 2
c 1   
1
  2 x 
2
2 2 1
 c
x  1  1
when t = 0, x = 1  2 
 x 
2
i.e. 0  2  c t 2
x
c 2
 t  2 
2 2
2
t  2 x
x
2
x
t  2 2

41. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1


42. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 

43. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2

44. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5

45. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5
1 1
 25  16    4   c
2 2
1
c
2

46. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5
1 1
 25  16    4   c
2 2
1
c
2
v2  x4  2x2 1

47. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5
v  x  1
1 1
 25  16    4   c 2 2 2
2 2
1
c
2
v2  x4  2x2 1

48. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5
v  x  1
1 1
 25  16    4   c 2 2 2
2 2
1 v  x2 1
c
2
v2  x4  2x2 1

49. 2004 Extension 1 HSC Q5a)
A particle is moving along the x axis starting from a position 2 metres to
the right of the origin (that is, x = 2 when t = 0) with an initial velocity
of 5 m/s and an acceleration given by   2 x 3  2 x
x
(i) Show that x  x 2  1

d 1 2 
 v   2x  2x
3
dx  2 
1 2 1 4
v  x  x2  c
2 2
When x = 2, v = 5
v  x  1
1 1
 25  16    4   c 2 2 2
2 2
1 v  x2 1
c
2 Note: v > 0, in order
v2  x4  2x2 1 to satisfy initial
conditions

50. (ii) Hence find an expression for x in terms of t

51. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt

52. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2

53. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x

54. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x
t  tan 1 x  tan 1 2

55. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x
t  tan 1 x  tan 1 2
tan 1 x  t  tan 1 2

56. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x
t  tan 1 x  tan 1 2
tan 1 x  t  tan 1 2
x  tan t  tan 1 2 

57. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x
t  tan 1 x  tan 1 2
tan 1 x  t  tan 1 2
x  tan t  tan 1 2 
tan t  2
x
1  2 tan t

58. (ii) Hence find an expression for x in terms of t
dx
 x2 1
dt
t x
dx
 dt   x 2  1
0 2
t  tan x 2
1 x
Exercise 3E; 1 to 3 acfh,
t  tan 1 x  tan 1 2 7 , 9, 11, 13, 15, 17, 18,
20, 21, 24*
tan 1 x  t  tan 1 2
x  tan t  tan 1 2 
tan t  2
x
1  2 tan t