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*