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Mechanics, EXT2 M1 2025 HSC 8 MC

The graph shows the velocity of a particle as a function of its displacement.
 

  

Which of the following graphs best shows the acceleration of the particle as a function of its displacement?
 

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\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(v(x)\ \Rightarrow\ \text{degree 3 (see graph)},\ \ \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 2}\)

\(a=v \cdot \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 5 (eliminate C and D)}\)

\(\text{At}\ \ x=0,\ \ v(x)>0\ \ \text{and}\ \ \dfrac{dv}{dx}<0\ \ \Rightarrow\ \ v \cdot \dfrac{dv}{dx} \neq 0\ \text{(eliminate B)}\)

\(\Rightarrow A\)

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 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 2020 SPEC2 17 MC

The velocity, `v` ms`\ ^(−1)`, of a particle at time  `t >= 0`  seconds and at position  `x >= 1`  metre from the origin is  `v = 1/x`.

The acceleration of the particle, in `text(ms)^(−2)`, when  `x = 2`  is

  1. `−1/4`
  2. `−1/8`
  3. `1/8`
  4. `1/4`
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`B`

Show Worked Solution

`v = 1/x`

`a` `= v · (dv)/(dx)`
  `= 1/x · −1/(x^2)`
  `= −1/(x^3)`

 

`text(When)\ \ x = 2:`

`a = −1/(2^3) = −1/8`

`=>B`

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 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 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 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 2016 HSC 15b

A particle is initially at rest at the point `B` which is `b` metres to the right of `O.`

The particle then moves in a straight line towards `O.`

For `x != 0,` the acceleration of the particle is given by  `(- mu^2)/x^2,`  where `x` is the distance from `O` and `mu` is a positive constant.

  1. Prove that  `(dx)/(dt) = -mu sqrt 2 sqrt((b - x)/(bx)).`  (2 marks)

  2. Using the substitution  `x = b cos^2 theta,` show that the time taken to reach a distance `d` metres to the right of `O` is given by
     
         `t = (b sqrt (2b))/mu int_0^(cos^-1 sqrt (d/b)) cos^2 theta\ d theta.`  (3 marks)

It can be shown that   `t = 1/mu sqrt (b/2) (sqrt(bd - d^2) + b cos^-1 sqrt (d/b)).`  (Do NOT prove this.)

  1. What is the limiting time taken for the particle to reach `O?`  (1 mark)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `(pi(sqrtb)^3)/(2sqrt2mu)\ text(seconds)`
Show Worked Solution

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

`:. 1/2 v^2` `= int −(mu^2)/(x^2)\ dx`
  `= (mu^2)/x + c`

 

`text(Initially,)\ v = 0\ text(and)\ x = b`

`:. c = −(mu^2)/b`

`v^2` `= 2mu^2(1/x – 1/b)`
`v` `= −musqrt2 · sqrt(1/x – 1/b)qquad(text(negative since moving to left))`
  `= −musqrt2 · sqrt((b – x)/(bx))\ …\ text(as required.)`

 

ii.    `dx/dt` `= −musqrt2 · sqrt((b – x)/(bx))`
  `dt/dx` `= −1/(musqrt2) · sqrt((bx)/(b – x))`
  `int_0^t dt` `= −1/(musqrt2) · int_b^d sqrt((bx)/(b – x))\ dx`

 

`text(Integration by substitution:)`

`text(Let)\ \ \ x` `= bcos^2theta`
`dx` `= −2bcosthetasintheta\ d theta`

 

`text(When)quadx` `= b,` `theta` `= 0`
`x` `= d,` `theta` `= cos^(−1)sqrt(d/b)`

 

`:. t` `= −1/(musqrt2) · int_0^(cos^(−1)sqrt(d/b))sqrt((b^2cos^2theta)/(b(1 – cos^2theta))) · −2bcosthetasintheta\ d theta`
 

`= (2b)/(musqrt2) · int_0^(cos^(−1)sqrt(d/b))(sqrtb costheta)/(sintheta) · costhetasintheta\ d theta`
 

`= (bsqrt(2b))/mu int_0^(cos^(−1)sqrt(d/b)) cos^2theta\ d theta\ …\ text(as required)`

 

iii.   `t = 1/mu sqrt(b/2)(sqrt(bd – d^2) + bcos^(−1)sqrt(d/b))`

 

`text(As)\ \ d->0,`

♦♦ Mean mark 29%.
`t` `= 1/mu sqrt(b/2) (sqrt0 + bcos^(−1)0)`
  `= 1/mu sqrt(b/2) · b · pi/2`
  `= (pi(sqrtb)^3)/(2sqrt2mu)\ text(seconds)`

Mechanics, EXT2 M1 2006 HSC 6b

In an alien universe, the gravitational attraction between two bodies is proportional to  `x^(–3)`, where  `x`  is the distance between their centres.

A particle is projected upward from the surface of a planet with velocity  `u`  at time  `t = 0`.  Its distance  `x`  from the centre of the planet satisfies the equation

`ddot x = - k/x^3.`

  1. Show that  `k =gR^3`, where  `g`  is the magnitude of the acceleration due to gravity at the surface of the planet and  `R`  is the radius of the planet.  (1 mark)

  2. Show that  `v`, the velocity of the particle, is given by
     
         `v^2 = (gR^3)/x^2 - (gR - u^2).`  (3 marks)

  3. It can be shown that  `x = sqrt (R^2 + 2uRt - (gR - u^2) t^2).` (Do NOT prove this.)

     

    Show that if  `u >= sqrt (gR)`  the particle will not return to the planet.  (2 marks)

  4. If  `u < sqrt (gR)`  the particle reaches a point whose distance from the centre of the planet is  `D`, and then falls back.

  5.  

      (1)   Use the formula in part (ii) to find  `D`  in terms of  `u, R`  and  `g.`  (1 mark)

  6.  

      (2)   Use the formula in part (iii) to find the time taken for the particle to return to the surface of the planet in terms of  `u, R`  and  `g.`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(1)\ \ D = sqrt ((gR^3)/(gR – u^2))`

     

    `(2)\ \ (2uR)/(gR – u^2)`

Show Worked Solution
i.   

`text(When)\ x=R,\ \ ddot x = – k/(R^3)`

`text(Given)\ \ ddot x = -g\ \ text(on the surface)`

`-g ­=` `-k/R^3`
`:.k ­=` `gR^3`

 

ii.    `ddot x` `=- (gR^3)/x^3`
  `1/2 v^2` `= int – (gR^3)/x^3\ dx`
    `= (gR^3)/(2x^2)+c_1`
  `:.v^2` `=(gR^3)/x^2 +c_2`

 
`text(When)\ t=0, x=R and v=u`

`u^2` `=(gR^3)/R^2 +c_2`
`c_2` `=u^2-gR`
`:.v^2` `=(gR^3)/x^2 +u^2-gR`
  `=(gR^3)/x^2 -(gR-u^2)`

 

iii. `text(Solution 1)`

`text(If)\ \ u >= sqrt (gR)\ \ text(then)\ \ u^2 >= gR`

`x` `= sqrt (R^2 + 2uRt – (gR – u^2) t^2)`
  `≥sqrt (R^2 + 2sqrt(gR)Rt – (gR – gR) t^2)`
  `≥sqrt (R^2 + 2sqrt(gR)Rt)`
  `>sqrt (R^2)\ \ \ \ (t>0)`
  `>R`

 

`:. x>R\ \ text(when)\ \ t>0,\ text(and the particle does not)`

`text(return to the surface of the planet.)`

 

`text(Solution 2)`

`v^2` `=(gR^3)/x^2 -(gR-u^2)`
  `>=(gR^3)/x^2\ \ \ \ text{(since}\ u^2 >= gR text{)}`
`v` `>=0`

 
`:.\ text(S)text(ince)\ \ v>=0,\ \ text(the particle is never moving back)`

`text(towards the planet and will never return.)`

 

iv. (1)  `v^2 = (gR^3)/D^2 – (gR – u^2)`

`x = D\ \ text(occurs when)\ \ v = 0`

`:.0 ­=` `(gR^3)/D^2 – (gR – u^2)`
`D^2 ­=` `(gR^3)/(gR – u^2)`
`:.D ­=` `sqrt ((gR^3)/(gR – u^2))`

 

iv. (2)  `text(Find)\ \ t\ \ text(when)\ \ x = R`

`text{Using part (iii)}`

`R` `=sqrt (R^2 + 2uRt – (gR – u^2) t^2)`
`R^2` `= R^2 + 2uRt – (gR – u^2) t^2`
`0` `= t(2uR – (gR – u^2)t)`
`:.t` ` = (2uR)/(gR – u^2)\ \ \ \ (t>0)`

 
`:.\ text(It takes)\ \ (2uR)/(gR – u^2)\ \ text(seconds to return to the planet.)`

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)

Show Answers Only
  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 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)

Show Answers Only
  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)

Show Answers Only
  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`

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`