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Mechanics, EXT2 M1 2025 HSC 14b

The acceleration of a particle is given by  \(\ddot{x}=32 x\left(x^2+3\right)\), where \(x\) is the displacement of the particle from a fixed-point \(O\) after \(t\) seconds, in metres. Initially the particle is at \(O\) and has a velocity of 12 m s\(^{-1}\)  in the negative direction.

  1. Show that the velocity of the particle is given by  \(v=-4\left(x^2+3\right)\).   (2 marks)

  2. Find the time taken for the particle to travel 3 metres from the origin.   (2 marks)

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i.    \(\text{See Worked Solutions}\)

ii.   \(t=\dfrac{\pi}{12 \sqrt{3}} \ \text{sec}\)

Show Worked Solution

i.    \(\ddot{x}=32 x\left(x^2+3\right)\)

\(\text{Show} \ \ v=-4\left(x^2+3\right)\)

\(\text{Using} \ \ \ddot{x}=v \cdot \dfrac{dv}{dx}:\)

\(v \cdot \dfrac{dv}{dx}\) \(=32 x\left(x^2+3\right)\)
\(\displaystyle \int v \, dv\) \(=\displaystyle \int 32 x^3+96 x\, dx\)
\(\dfrac{v^2}{2}\) \(=8 x^4+48 x^2+c\)

 
\(\text{When} \ \ x=0, v=-12 \ \Rightarrow \ c=72\)

\(\dfrac{v^2}{2}=8 x^4+48 x^2+72\)

\(v^2=16\left(x^4+6 x^2+9\right)\)

\(v=-4\left(x^2+3\right) \quad (V=-12 \ \ \text {when} \ \ x=0)\)
 

ii.   \(\dfrac{dx}{dt}=-4\left(x^2+3\right)\)

\(\dfrac{dt}{dx}=-\dfrac{1}{4\left(x^2+3\right)}\)

\(t=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{3+x^2} d x=-\frac{1}{4} \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c\)

\(\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=0\)
 

\(\text{Since particle is moving left at} \ \ t=0,\)

\(\text{Find \(t\) when} \ \ x=-3:\)

\(t\) \(=-\dfrac{1}{4 \sqrt{3}} \times \tan ^{-1}\left(-\dfrac{3}{\sqrt{3}}\right)\)
  \(=-\dfrac{1}{4 \sqrt{3}} \times \tan ^{-1}(-\sqrt{3})\)
  \(=-\dfrac{1}{4 \sqrt{3}} \times-\dfrac{\pi}{3}\)
  \(=\dfrac{\pi}{12 \sqrt{3}} \ \text{sec}\)

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\ text{ms}^{-2}` of a body moving in a straight line in terms of the velocity `v\ text{ms}^{-2}` 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\ \ =>\ \ c1-1/4 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`?

  1. `2\ text(m s)^(-2)`
  2. `3\ text(m s)^(-2)`
  3. `6\ text(m s)^(-2)`
  4. `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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a.    `text(Show Worked Solutions)`

b.    `tan\ (t + tan^(−1)2)`

Show Worked Solution

a.    `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`
`1/2 v^2` `= 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\ \ =>\ \ 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)`

  

b.     `(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
  3.    `int 1/(x(1 + x)) \ dx = -3t.`   (2 marks)

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

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `x = 2/(e^(3t)-2)`

Show Worked Solution

a.    `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`
`1/2 v^2` `= 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)`

 

b.    `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)`

 

c.    `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`

 

`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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a.    `text(See Worked Solutions.)`

b.    `x = 1-e^(-t)`

c.    `x = 1`

Show Worked Solution

a.    `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)`

 

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

 

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

♦ Mean mark (b) 49%.

`0= -ln1 + c\ \ =>\ \ c=0`

`t` `= -ln(1-x)`
`-t` `= ln(1-x)`
`1-x` `= e^(-t)`
`:.x` `= 1-e^(-t)`

 

♦ Mean mark (c) 2%.

c.    `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 1/2`

`:.\ 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`