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
2
dx  2 
dt

4. Velocity & Acceleration in
Terms of x
If v = f(x);
d 2x d  1 2 
=  v
2
dx  2 
dt
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
2
dx  2 
dt
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
2
dx  2 
dt
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
2
dx  2 
dt
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
2
dx  2 
dt
Proof: d 2 x dv
=
2
dt dt
dv dx
=⋅
dx dt
dv
= ⋅v
dx
dv d  1 2 
= ⋅  v
dx dv  2 
=  v2 
d1
 
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 
12
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 
12
v = 3x − x 2 + c
2
when x = 1, v = 2
12
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 
12
v = 3x − x 2 + c
2
when x = 1, v = 2
12
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 
12
v = 3x − x 2 + c
2
when x = 1, v = 2
12
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:
12
v = 3x − x 2 + c
v2 ≥ 0
2
when x = 1, v = 2
12
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:
12
v = 3x − x 2 + c
v2 ≥ 0
2
when x = 1, v = 2 6x − 2x2 ≥ 0
12
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:
12
v = 3x − x 2 + c
v2 ≥ 0
2
when x = 1, v = 2 6x − 2x2 ≥ 0
12 2 x( 3 − x ) ≥ 0
i.e. ( 2 ) = 3(1) − 12 + c
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:
12
v = 3x − x 2 + c
v2 ≥ 0
2
when x = 1, v = 2 6x − 2x2 ≥ 0
12 2 x( 3 − x ) ≥ 0
i.e. ( 2 ) = 3(1) − 12 + c
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:
12
v = 3x − x 2 + c
v2 ≥ 0
2
when x = 1, v = 2 6x − 2x2 ≥ 0
12 2 x( 3 − x ) ≥ 0
i.e. ( 2 ) = 3(1) − 12 + c
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/sin
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/sin
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/sin
the negative direction. Find x in terms of t.
d 1 2 
 v  = 3x
2
dx  2 
12
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/sin
the negative direction. Find x in terms of t.
d 1 2 
 v  = 3x
2
dx  2 
12
v = x3 + c
2
when t = 0, x = 1, v = − 2
1
( − 2 ) 2 = 13 + c
i.e.
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/sin
the negative direction. Find x in terms of t.
d 1 2 
 v  = 3x
2
dx  2 
12
v = x3 + c
2
when t = 0, x = 1, v = − 2
1
( − 2 ) 2 = 13 + c
i.e.
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/sin
the negative direction. Find x in terms of t.
d 1 2  dx
 v  = 3x
2
= − 2x 3
dx  2  dt
12
v = x3 + c
2
when t = 0, x = 1, v = − 2
1
( − 2 ) 2 = 13 + c
i.e.
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/sin
the negative direction. Find x in terms of t.
d 1 2  3 (Choose –ve to satisfy
dx
 v  = 3x
2
= − 2x the initial conditions)
dx  2  dt
12
v = x3 + c
2
when t = 0, x = 1, v = − 2
1
( − 2 ) 2 = 13 + c
i.e.
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/sin
the negative direction. Find x in terms of t.
d 1 2  3 (Choose –ve to satisfy
dx
 v  = 3x
2
= − 2x the initial conditions)
dx  2  dt
12 3
v = x3 + c
= − 2x 2
2
when t = 0, x = 1, v = − 2
1
( − 2 ) 2 = 13 + c
i.e.
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/sin
the negative direction. Find x in terms of t.
d 1 2  3 (Choose –ve to satisfy
dx
 v  = 3x
2
= − 2x the initial conditions)
dx  2  dt
12 3
v = x3 + c
= − 2x 2
2
when t = 0, x = 1, v = − 2 3
dt 1 −2
=−
1
( − 2 ) 2 = 13 + c x
i.e. dx 2
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 +c
2
2
= +c
x

31. 3
dt 1 −2
=− x
dx 2
1
1 −
t=− ⋅ −2 x 2 + c
2
1
−
= 2x +c
2
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 +c
2
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 +c
2
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 +c
2
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 t=− x 2 dx
OR
2 1
1
−
= 2x +c
2
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 t=− x 2 dx
OR
2 1
1 x
−
1  1
= 2x +c −
2
=− − 2 x 
2
2 1
2
= +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 t=− x 2 dx
OR
2 1
1 x
−
1  1
= 2x +c −
2
=− − 2 x 
2
2 1
2
= +c
x  1 − 1
= 2 
when t = 0, x = 1
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 t=− x 2 dx
OR
2 1
1 x
−
1  1
= 2x +c −
2
=− − 2 x 
2
2 1
2
= +c
x  1 − 1
= 2 
when t = 0, x = 1
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 t=− x 2 dx
OR
2 1
1 x
−
1  1
= 2x +c −
2
=− − 2 x 
2
2 1
2
= +c
x  1 − 1
= 2 
when t = 0, x = 1
x 
2
i.e. 0 = 2 + c t+ 2=
x
c=− 2
2
(t + 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 t=− x 2 dx
OR
2 1
1 x
−
1  1
= 2x +c −
2
=− − 2 x 
2
2 1
2
= +c
x  1 − 1
= 2 
when t = 0, x = 1
x 
2
i.e. 0 = 2 + c t+ 2=
x
c=− 2
2
(t + 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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
v = x + x2 + c
2 2
When x = 2, v = 5
1 1
v = ( x + 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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
v = x + x2 + c
2 2
When x = 2, v = 5
1 1
v = ( x + 1)
( 25) = ( 16 ) + ( 4 ) + c 2
2 2
2 2
v = x2 +1
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
 = 2 x 3 + 2 x
x
5 m/s and an acceleration given by
(i) Show that x = x 2 + 1

d 1 2 
 v  = 2x + 2x
3
dx  2 
12 14
v = x + x2 + c
2 2
When x = 2, v = 5
1 1
v = ( x + 1)
( 25) = ( 16 ) + ( 4 ) + c 2
2 2
2 2
v = x2 +1
1
c=
2 Note: v > 0, in order
to satisfy initial
v2 = x4 + 2x2 +1
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
x
−1

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
x
−1
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
x
−1
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
x
−1
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
x
−1
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
x
−1
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