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Mechanics, EXT2* M1 2019 HSC 13c

A particle moves in a straight line. At time  `t`  seconds the particle has a displacement of `x` m, a velocity of  `v\ text(m s)^(-1)`  and acceleration  `a\ text(m s)^(-2)`.

Initially the particle has displacement  0 m  and velocity  `2\ text(m s)^(-1)`. The acceleration is given by  `a = -2e^(-x)`. The velocity of the particle is always positive.

  1. Show that  `v = 2e^((-x)/2)`.  (2 marks)

  2. Find an expression for `x` as a function of  `t`.   (2 marks)

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  1. `text(See Worked Solutions)`
  2. `x = 2 ln(t + 1)`
Show Worked Solution
i.    `a` `= -2e^(-x)`
  `d/(dx)(1/2v^2)` `= -2e^(-x)`
  `1/2 v^2` `= int -2e^(-x) dx`
    `= 2e^(-x) + C`

 
`text(When)\ \ x = 0,\ \ v = 2:`

`1/2 ⋅ 2^2` `= 2e^0 + C`
`C` `= 0`
`1/2 v^2` `= 2e^(-x)`
`v^2` `= 4e^(-x)`
`v` `= +-(4e^(-x))^(1/2)`
  `= +-2e^(-x/2)`

 
`text(S)text(ince)\ \ v = 2\ \ text(when)\ \ x = 0,`

`v = 2e^((-x)/2)`

 

ii.    `(dx)/(dt)` `= 2e^((-x)/2)`
  `(dt)/(dx)` `= (e^(x/2))/2`
  `t` `= 1/2 int e^(x/2) dx`
    `= 1/2 xx 2 xx e^(x/2) + C`
    `= e^(x/2) + C`

 
`text(When)\ \ t = 0,\ \ x = 0:`

♦ Mean mark part (ii) 46%.

`0` `= e^0 + C`
`C` `= -1`
`t` `= e^(x/2) – 1`
`e^(x/2)` `= t + 1`
`x/2` `= ln (t + 1)`
`:. x` `= 2 ln (t + 1)`

Mechanics, EXT2* M1 2017 HSC 12d

At time `t` the displacement, `x`, of a particle satisfies  `t=4-e^(-2x)`.

Find the acceleration of the particle as a function of `x`.  (3 marks)

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`(e^(4x))/2`

Show Worked Solution
`t` `= 4 – e^(−2x)`
`(dt)/(dx)` `= 2e^(−2x)`
`(dx)/(dt)` `= (e^(2x))/2`

 

`a` `= {:d/(dx):}^(1/2v^2)`
  `= d/(dx)(1/2 · ((e^(2x))/2)^2)`
  `= d/(dx)((e^(4x))/8)`
  `= 4 · (e^(4x))/8`
  `= (e^(4x))/2`

Mechanics, EXT2* M1 2007 HSC 3c

A particle is moving in a straight line with its acceleration as a function of  `x`  given by  `ddot x = -e^(-2x)`. It is initially at the origin and is travelling with a velocity of 1 metre per second.

  1. Show that  `dot x = e^(-x)`.  (2 marks)

  2. Hence show that  `x = log_e(t + 1)`.  (2 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `text(Show that)\ \ dot x = e^(−x)`

`ddot x` `= d/(dx)\ (1/2 (dot x)^2) = −e^(−2x)`
`1/2 (dot x)^2` `= int −e^(−2x)\ dx`
  `= 1/2 e^(−2x) + c`

 
`text(When)\ \ x = 0, \ dot x = 1,`

`1/2 · 1^2` `= 1/2 e^0 + c`
`1/2` `= 1/2 + c`
`:.c` `= 0`

 

MARKER’S COMMENT: Most candidates “neglected” to consider the two cases that  `dot x=+-e^(-x)`.
`1/2 (dot x)^2` `= 1/2 e^(−2x)`
`(dot x)^2` `= e^(−2x)`
`dot x` `= +-e^(−x)`

 
`text(Given initial conditions:)\ \ x=0, dot x = 1,`

`dot x = e^(-x)\ \ text(… as required)`

 

ii.   `text(Show)\ \ x = log_e(t + 1)`

`(dx)/(dt)` `= e^(−x)`
`(dt)/(dx)` `= e^x`
`t` `= int e^x`
  `= e^x + c`

 

`text(When)\ \ t = 0, \ x = 0`

`0` `= e^0 + c`
`c` `= −1`
`:.t` `= e^x − 1`
`e^x` `= t + 1`
`log_ee^x` `= log_e(t + 1)`
`x` `= log_e(t + 1)\ …\ text(as required.)`

Mechanics, EXT2* M1 2007 HSC 2d

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling  `t`  seconds after jumping is given by  `v =50(1 - e^(-0.2t))`.

  1. Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.  (2 marks)

  2. Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.  (2 marks)

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  1. `1.4\ text(ms)^(−2)\ \ text{(to 1 d.p.)}`
  2. `284\ text{m  (nearest m)}`
Show Worked Solution
i.     `v` `= 50(1 − e^(-0.2t))`
    `=50-50e^(-0.2t)`
  `ddot x` `= (dv)/(dt)`
    `= −0.2 xx 50 xx −e^(−0.2t)`
    `= 10 e^(−0.2t)`

 
`text(When)\ \ t = 10:`

`ddot x` `= 10 e^(−0.2 xx 10)`
  `= 10 e ^(−2)`
  `= 1.353…`
  `= 1.4\ text(ms)^(−2)\ \ \ text{(to 1 d.p.)}`

 

ii.  `text(Distance travelled)`

`= int_0^10 v\ dt`

`= 50 int_0^10 1 − e^(−0.2t) \ dt`

`= 50 [t + 1/0.2 · e^(−0.2t)]_0^10`

`= 50 [t + 5e^(−0.2t)]_0^10`

`= 50 [(10 + 5e^(−2)) − (0 + 5e^0)]`

`= 50 [10 + 5e^(−2) − 5]`

`= 50 [5 + 5e^(−2)]`

`= 283.833…`

`= 284\ text{m  (nearest m)}`

Mechanics, EXT2* M1 2014 HSC 12c

A particle moves along a straight line with displacement  `x\ text(m)`  and velocity  `v\ \ text(ms)^(-1)`. The acceleration of the particle is given by

 `ddot x = 2 - e^(-x/2)`.

Given that  `v = 4`  when  `x = 0`, express  `v^2`  in terms of  `x`.   (3 marks)

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`v^2 = 4x + 4e^(-x/2) + 12`

Show Worked Solution

`ddot x = d/(dx) (1/2 v^2) = 2 – e^(-x/2)`

`1/2 v^2` `= int (2 – e^(-x/2))\ dx`
  `= 2x + 2e^(-x/2) + c`
`v^2` `= 4x + 4e^(-x/2) + c`

 
`text(When)\ \ v = 4, x = 0:`

`:. 4^2` `= 0 + 4e^0 + c`
`16` `= 4 + c`
`c` `= 12`

 
`:.\ v^2 = 4x + 4e^(-x/2) + 12`