The document discusses velocity and acceleration in terms of position x. It provides equations showing that acceleration is equal to the derivative of velocity with respect to time, and the derivative of velocity with respect to x. It also gives examples of using these relationships to find velocity and position as functions of x and time for particles where acceleration is given.

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