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Mechanics, EXT2 M1 2023 SPEC1 8

A body moves in a straight line so that when its displacement from a fixed origin `O` is `x` metres, its acceleration, `a`, is `-4 x \ text{ms}^{-2}`. The body accelerates from rest and its velocity, `v`, is equal to `-2 \ text{ms}^{-1}` as it passes through the origin. The body then comes to rest again.

Find `v` in terms of `x` for this interval.  (4 marks)

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` – 2sqrt(1-x^2)`

Show Worked Solution

`a = -4x`

`d/dx(1/2v^2)` `= -4x`  
`1/2v^2` `= -2x^2 +c`  

 
`v= -2\ \ \text{when}\ \ x = 0\ \ =>\ \ c = 2`

`v^2` `= -4x^2 + 4`  
`v^2` `= 4(1-x^2)`  

 
`v= -2\ \ text{when}\ \ x=0:`

`:.\ v` `= -sqrt(4(1-x^2))`  
  `= -2sqrt(1-x^2)`  

Mechanics, EXT2 M1 2017 SPEC2 17 MC

The acceleration,  `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by  `a = v^2 + 1`,  where  `v`  is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin O is  `2\ text(ms)^(-1)`.

The displacement of the particle from O when its velocity is  `3\ text(ms)^(-1)`  is

  1. `log_e(2)`
  2. `1/2 log_e(10/3)`
  3. `1/2 log_e(2)`
  4. `1/2 log_e(5/2)`
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`C`

Show Worked Solution
`v* (dv)/(dx)` `= v^2 + 1`
`(dv)/(dx)` `= (v^2 + 1)/v`
`(dx)/(dv)` `= v/(v^2 + 1)`
`x` `= int (v/(v^2 + 1))\ dv`
  `=1/2 ln(v^2 + 1)+c`

 
`text(When)\ \ x=0, \ v=2:`

`c=-1/2 ln5`

 
`text(Find)\ \ x\ \ text(when)\ \ v=3:`

`x` `= 1/2ln(9 + 1) – 1/2 ln5`
  `= 1/2 ln (10/5)`
  `= 1/2 ln 2`

 
`=>   C`

Mechanics, EXT2 M1 2015 SPEC1 6

The acceleration `a` ms¯² of a body moving in a straight line in terms of the velocity `v` ms¯¹ is given by  `a = 4v^2.`

Given that  `v = e`  when  `x = 1`, where `x` is the displacement of the body in metres, find the velocity of the body when  `x = 2.`  (4 marks)

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`e^5`

Show Worked Solution
`v* (dv)/(dx)` `= 4v^2`
`(dv)/(dx)` `= 4v`
`(dx)/(dv)` `= 1/(4v)`
`x` `=int 1/(4v) \ dv`
  `=1/4 ln (4v)+c`

 
`text(When)\ \ x=1, v=e`

`1` `=1/4 ln(4e)+c`  
`:.c` `=1-1/4 ln(4e)`  

 
`text(Find)\ \ v\ \ text(when)\ \ x=2:`

COMMENT: Strong log and exponential calculation ability required here.

`1/4 ln (4v)+1-1/4 ln(4e)` `= 2`
`1/4 ln(4v)` `= 1/4 ln (4e)+1`
`ln(4v)` `= ln(4e)+4`
`e^(ln4v)` `= e^(ln(4e) +4)`
`4v` `= e^(ln(4e)) * e^4`
`4v` `=4e xx e^4`
`:.v` `=e^5`

Mechanics, EXT2 M1 2020 HSC 11c

A particle starts at the origin with velocity 1 and acceleration given by

`a = v^2 + v`,

where `v` is the velocity of the particle.

Find an expression for `x`, the displacement of the particle, in terms of `v`.   (3 marks)

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`x= ln \ | frac{v + 1}{2} |`

Show Worked Solution
`a` `= v^2 + v`
`v · frac{dv}{dx}` `= v^2 + v`
`frac{dv}{dx}` `= v + 1`
`frac{dx}{dv}` `= frac{1}{v + 1}`
`x` `= int frac{1}{v + 1}\ dv`
  `= ln \ | v + 1 | + c`

 
`text{When} \ \ x = 0 , \ v = 1`

`0 = ln \ 2 + c`

`c = -ln \ 2`

`therefore \ x` `= ln \ | v + 1| – ln \ 2`
  `= ln \ | frac{v + 1}{2} |`

Mechanics, EXT2* M1 2018 HSC 7 MC

The velocity of a particle, in metres per second, is given by  `v = x^2 + 2`  where `x` is its displacement in metres from the origin.

What is the acceleration of the particle at  `x = 1`?

A.     `2\ text(m s)^(-2)`

B.     `3\ text(m s)^(-2)`

C.     `6\ text(m s)^(-2)`

D.     `12\ text(m s)^(-2)`

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`C`

Show Worked Solution
`a` `= d/dx (1/2  v^2)`
  `= d/dx (1/2 (x^2 + 2)^2)`
  `=1/2 xx 2 xx 2x (x^2+2)`
  `= 2x (x^2 + 2)`

 
`text(When)\ \ x = 1,`

`a = 6\ text(m s)^(-2)`
 

`⇒  C`

Mechanics, EXT2* M1 2004 HSC 5a

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\ text(ms)^(−1)`  and an acceleration given by

`ddot x = 2x^3 + 2x`.

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

  2. Hence find an expression for  `x`  in terms of  `t`.  (3 marks)

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  1. `text(Show Worked Solutions)`
  2. `tan\ (t + tan^(−1)2)`
Show Worked Solution

i.   `text(Show)\ \ dot x = x^2 + 1`

`ddot x` `= d/(dx)\ (1/2 v^2) = 2x^3 + 2x`
`:.1/2 v^2` `= int2x^3 + 2x \ dx`
  `= 2/4x^4 + x^2 + c`
`v^2` `= x^4 + 2x^2 + c`

 
`text(When)\ \ x = 2, \ v = 5`

`5^2` `= 2^4 + (2 xx 2^2) + c`
`25` `= 16 + 8 + c`
`c` `= 1`
`:.v^2` `= x^4 + 2x^2 + 1`
  `= (x^2 + 1)^2`
`v` `= sqrt((x^2 + 1)^2)`
 `:.dot x` `= x^2 + 1\ \ \ …\ text(as required)`

  

ii.    `(dx)/(dt)` `= x^2 + 1`
  `(dt)/(dx)` `= 1/(x^2 + 1)`
  `:.t` `= int1/(x^2 + 1)\ dx`
    `= tan^(−1)\ x + c`

 
`text(When)\ \ t = 0, \ x = 2`

`0` `= tan^(−1)\ 2 + c`
`c` `= −tan^(−1)\ 2`
`:.t` `=tan^(−1)\ x − tan^(−1)\ 2`

 

`tan^(−1)\ x` `= t + tan^(−1)2`
`:.x` `= tan (t + tan^(−1)2)`

Mechanics, EXT2* M1 2006 HSC 4c

A particle is moving so that  `ddot x = 18x^3 + 27x^2 + 9x.`

Initially  `x = – 2`  and the velocity, `v`, is  `– 6.`

  1. Show that  `v^2 = 9x^2 (1 + x)^2.`  (2 marks)

  2. Hence, or otherwise, show that 
     
         `int 1/(x(1 + x)) \ dx = -3t.`  (2 marks)

  3. It can be shown that for some constant  `c`,
     
         `log_e (1 + 1/x) = 3t + c.`       (Do NOT prove this.)
     
    Using this equation and the initial conditions, find  `x`  as a function of  `t.`  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `x = 2/(e^(3t) – 2)`
Show Worked Solution

i.   `text(Show)\ \ v^2 = 9x^2 (1 + x)^2`

`ddot x` `= d/(dx) (1/2 v^2)`
  `= 18x^3 + 27x^2 + 9x`
`1/2 v^2` `= int 18x^3 + 27x^2 + 9x\ dx`
  `= 18/4 x^4 + 27/3 x^3 + 9/2 x^2 + c`
`v^2` `= 9x^4 + 18x^3 + 9x^2 + c`

 

`text(When)\ \ t = 0,\ \ x = -2,\ \ v = -6`

`:.\ (-6)^2` `= 9 (-2)^4 + 18 (-2)^3 + 9 (-2)^2 + c`
`36` `= 144 – 144 + 36 + c`
`c` `= 0`

 

`:.\ v^2` `= 9x^4 + 18x^3 + 9x^2`
  `= 9x^2 (x^2 + 2x + 1)`
  `= 9x^2 (1 + x)^2\ \ text(…  as required)`

 

ii.   `text(Show)\ \ int 1/(x (1 + x)) \ dx = -3t`

`v^2 = 9x^2 (1 + x)^2`

`v = +- sqrt (9x^2 (1 + x)^2)`
 

`text(When)\ \ x = -2,\ \ v = -6:`

`:.\ v` `= -sqrt (9x^2 (1 + x)^2)`
  `= -3x (1 + x)`

 

`(dx)/(dt)` `= -3x (1 + x)`
`(dt)/(dx)` `= -1/(3x (1 + x))`
`t` `= -1/3 int 1/(x (1 + x)) \ dx`
`-3t` `= int 1/(x (1 + x)) \ dx\ \ text(…  as required)`

 

iii.  `text(Given)\ \ log_e (1 + 1/x) = 3t + c`

`text(When)\ \ t = 0,\ \ x = -2:`

`log_e (1 – 1/2)` `= 3(0) + c`
`log_e (1/2)` `= c`
`c` `= log_e 2^-1`
  `= -log_e 2`

 

`log_e (1 + 1/x)` `= 3t – log_e 2`
`1 + 1/x` `= e^(3t – log_e 2)`
`1/x` `= e^(3t – log_e 2) – 1`
  `= (e^(3t))/(e^(log_e 2)) – 1`
  `= e^(3t)/2 – 1`
  `= (e^(3t) – 2)/2`
`:.\ x` `= 2/(e^(3t) – 2)`

Mechanics, EXT2* M1 2015 HSC 14b

A particle is moving horizontally. Initially the particle is at the origin `O` moving with velocity `1 text(ms)^(−1)`.

The acceleration of the particle is given by  `ddot x = x − 1`, where `x` is its displacement at time  `t`.

  1. Show that the velocity of the particle is given by  `dot x = 1 − x`.  (3 marks)

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

  3. Find the limiting position of the particle.  (1 mark)

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  1. `text(See Worked Solutions.)`
  2. `x = 1 − e^(-t)`
  3. `x = 1`
Show Worked Solution

i.   `ddot x = d/(dx)(1/2 v^2) = x − 1`

`1/2 v^2` `= int ddot x\ dx`
  `= int x − 1\ dx`
  `= 1/2x^2 − x + c`

 
`text(When)\ \ x = 0, \ v = 1:`

`1/2·1^2` `= 0 − 0 + c`
`c` `= 1/2`

 

`:.1/2  v^2` `= 1/2x^2 − x + 1/2`
`v^2` `= x^2 − 2x + 1`
  `= (x − 1)^2`
`:.dot x` `= ±(x − 1)`

 
`text(S)text(ince)\ \ dotx = 1\ \ text(when)\ \ x = 0,`

`dot x = 1 − x\ \ …\ text(as required)`

 

ii.   `(dx)/(dt)` `= 1 − x\ \ \ text{(from (i))}`
  `(dt)/(dx)` `= 1/(1 − x)`
  `t` `= int 1/(1 − x)\ dx`
    `= -ln(1 − x) + c`

 

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

♦ Mean mark 49%.
`0` `= -ln1 + c`
`:.c` `= 0`
`t` `= -ln(1 − x)`
`-t` `= ln(1 − x)`
`1 − x` `= e^(-t)`
`:.x` `= 1 − e^(-t)`

 

♦ Mean mark 42%.
iii.   `text(As)\ t` `→ ∞`
  `e^(-t)` `→ 0`
  `x` `→ 1`

 

`:.\ text(Limiting position is)\ \ x = 1`

Mechanics, EXT2* M1 2008 HSC 2b

A particle moves on the  `x`-axis with velocity  `v`. The particle is initially at rest at  `x = 1`. Its acceleration is given by  `ddot x = x + 4`.

Using the fact that  `ddot x = d/dx (1/2 v^2)`, find the speed of the particle at  `x = 2`.   (3 marks)

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`sqrt11`

Show Worked Solution
`ddot x` `= d/dx (1/2 v^2)=x+4`
`1/2 v^2` `= int ddot x\ dx`
  `= int x + 4\ dx`
  `= 1/2 x^2 + 4x + c`

 
`text(When)\ \ x = 1, v = 0`

`0` `= 1/2 + 4 + c`
`c` `= -4\ ½`

 

`:.\ 1/2 v^2` `= 1/2 x^2 + 4x – 4 1/2`
`v^2` `= x^2 + 8x – 9`

 
`text(When)\ \ x = 2`

`v^2` `= 2^2 + 8 * 2 – 9`
  `= 11`
`v` `= +- sqrt 11`

 
`:.\ text(When)\ x = 2,\ \ text(Speed) = sqrt 11`