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12 x1 t07 02 v and a in terms of x (2012)

N
Nigel Simmons

1) The document discusses relating acceleration to velocity and position for particles moving in one dimension. 2) It derives the formula for acceleration as the second derivative of position with respect to time, equal to the first derivative of velocity with respect to position. 3) It provides two examples: - Example 1 finds the velocity of a particle in terms of its position by solving the differential equation for acceleration. - Example 2 finds the position of a particle in terms of time by solving the differential equation and using initial conditions.

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Velocity & Acceleration in Terms of x
Velocity & Acceleration in Terms of x If v = f(x);
Velocity & Acceleration in Terms of x If v = f(x); d 2x d  1 2    v  dt 2 dx  2 
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
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
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
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 
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 
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.
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 
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
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
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
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
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
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
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
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
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
(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.
(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 
(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
(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
(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
(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
(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
(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
(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
3 dt 1 2  x dx 2
3 dt 1 2  x dx 2 1 1  t  2 x 2  c 2 1   2x 2 c 2  c x
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
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
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
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
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
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
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
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
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
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
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 
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 
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
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
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
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
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
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
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
(ii) Hence find an expression for x in terms of t
(ii) Hence find an expression for x in terms of t dx  x2 1 dt
(ii) Hence find an expression for x in terms of t dx  x2 1 dt t x dx  dt   x 2  1 0 2
(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
(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
(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
(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 
(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
(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

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12 x1 t07 02 v and a in terms of x (2012)

  • 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