The document discusses velocity and acceleration in terms of position (x). It defines acceleration as the derivative of velocity with respect to time and velocity as the derivative of position with respect to time. It then provides proofs of these relationships by taking derivatives. The document also includes an example problem that demonstrates finding the velocity of a particle moving in a straight line given its position function. It provides another example of finding the position of a particle over time given its initial position and velocity and its acceleration as a function of position.

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);





 2
2
2
2
1
v
dx
d
dt
xd

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

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

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

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

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

9. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23

10. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xv
dx
d
23
2
1 2






xx 23

11. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1

12. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx

13. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 

14. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 

15. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 
NOTE:
02
v

16. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 
NOTE:
02
v
026 2
 xx

17. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 
NOTE:
02
v
026 2
 xx
  032  xx

18. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 
NOTE:
02
v
026 2
 xx
  032  xx
30  x

19. e.g. (i) A particle moves in a straight line so that
Find its velocity in terms of x given that v = 2 when x = 1.
xx 23
xv
dx
d
23
2
1 2






cxxv  22
3
2
1
   
0
1132
2
1
i.e.
2,1when
22



c
c
vx
22
26 xxv 
2
26 xxv 
NOTE:
02
v
026 2
 xx
  032  xx
30  x
Particle moves between x = 0
and x = 3 and nowhere else.

20. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.

21. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d







22. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






cxv  32
2
1

23. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1

24. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1
3
32
2
2
xv
xv



25. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1
3
32
2
2
xv
xv


3
2x
dt
dx


26. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1
3
32
2
2
xv
xv


3
2x
dt
dx

(Choose –ve to satisfy
the initial conditions)

27. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1
3
32
2
2
xv
xv


3
2x
dt
dx

2
3
2x
(Choose –ve to satisfy
the initial conditions)

28. 2
3xx 
m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is
1 unit to the right of O, and is traveling with a velocity of in
the negative direction. Find x in terms of t.
22
3
2
1
xv
dx
d






 
0
12
2
1
i.e.
2,1,0when
32



c
c
vxt
cxv  32
2
1
3
32
2
2
xv
xv


3
2x
dt
dx

2
3
2x
(Choose –ve to satisfy
the initial conditions)
2
3
2
1 
 x
dx
dt

29. 2
3
2
1 
 x
dx
dt

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

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

32. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c

33. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t

34. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR

35. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1

36. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1
x
x
1
2
1
2
2
1









37. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1
x
x
1
2
1
2
2
1













  1
1
2
x

38. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1
x
x
1
2
1
2
2
1













  1
1
2
x
x
t
2
2 

39. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1
x
x
1
2
1
2
2
1













  1
1
2
x
x
t
2
2 
 2
2
2
 t
x

40. 2
3
2
1 
 x
dx
dt
c
x
cx
cxt





2
2
2
2
1
2
1
2
1
when t = 0, x = 1
2
20i.e.


c
c
2
2

x
t
OR 


x
dxxt
1
2
3
2
1
x
x
1
2
1
2
2
1













  1
1
2
x
x
t
2
2 
 2
2
2
 t
x
 2
2
2


t
x

41. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx

42. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32







43. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1

44. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5

45. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5
     
1 1
25 16 4
2 2
1
2
c
c
  


46. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5
     
1 1
25 16 4
2 2
1
2
c
c
  

12 242
 xxv

47. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5
     
1 1
25 16 4
2 2
1
2
c
c
  

12 242
 xxv
 222
1 xv

48. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5
     
1 1
25 16 4
2 2
1
2
c
c
  

12 242
 xxv
 222
1 xv
12
 xv

49. 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
2004 Extension 1 HSC Q5a)
xxx 22 3

(i) Show that 12
 xx
xxv
dx
d
22
2
1 32






cxxv  242
2
1
2
1
When x = 2, v = 5
     
1 1
25 16 4
2 2
1
2
c
c
  

12 242
 xxv
 222
1 xv
12
 xv
Note: v > 0, in order
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
12
 x
dt
dx

52. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1

53. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan


54. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan

2tantan 11 
 xt

55. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan

2tantan 11 
 xt
2tantan 11 
 tx

56. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan

2tantan 11 
 xt
2tantan 11 
 tx
 2tantan 1
 tx

57. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan

2tantan 11 
 xt
2tantan 11 
 tx
 2tantan 1
 tx
t
t
x
tan21
2tan




58. (ii) Hence find an expression for x in terms of t
12
 x
dt
dx
 

xt
x
dx
dt
2
2
0
1
 x
xt 2
1
tan

2tantan 11 
 xt
2tantan 11 
 tx
 2tantan 1
 tx
t
t
x
tan21
2tan



Exercise 3E; 1 to 3 acfh,
7 , 9, 11, 13, 15, 17, 18,
20, 21, 24*