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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

\(\text{By elimination:}\)

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

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

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

\(\Rightarrow A\)

♦♦ Mean mark 35%.

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)

Show Answers Only

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 2024 HSC 15c

A bar magnet is held vertically. An object that is repelled by the magnet is to be dropped from directly above the magnet and will maintain a vertical trajectory. Let \(x\) be the distance of the object above the magnet.
 

The object is subject to acceleration due to gravity, \(g\), and an acceleration due to the magnet \(\dfrac{27 g}{x^3}\), so that the total acceleration of the object is given by

 \(a=\dfrac{27 g}{x^3}-g\)

The object is released from rest at  \(x=6\).

  1. Show that  \(v^2=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)\).   (2 marks)

  2. Find where the object next comes to rest, giving your answer correct to 1 decimal place.   (2 marks)

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i.    \(\text{See Worked Solutions}\)

ii.   \(1.7 \ \text{units}\)

Show Worked Solution

i.    \(a=\dfrac{27 g}{x^3}-g\)

\(\dfrac{d}{dx}(\frac{1}{2}v^{2})\) \(= \dfrac{27g}{x^3}-g\)  
\(\dfrac{1}{2} v^2\) \(=-\dfrac{27 g}{2 x^2}-g x+c\)  

 
\(\text{When}\ \ x=6, v=0:\)

\(0\) \(=-\dfrac{27g}{2 \times 6^2}-6g+c\)  
\(c\) \(=\dfrac{459 g}{72}=\dfrac{51 g}{8}\)  

 

  \(\dfrac{1}{2} v^2\) \(=-\dfrac{27 g}{2 x^2}-g x+\dfrac{51 g}{8}\)
  \(v^2\) \(=-\dfrac{27 g}{x^2}-2 g x+\dfrac{51g }{4}\)
    \(=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)\)

  

ii.    \(\text{Find \(x\) when  \(v=0\):}\)

\(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\) \(=0\)  
\(51 x^2-8 x^3-108\) \(=0\)  
\(8 x^3-51 x^2+108\) \(=0\)  

 
\(\text{Given  \(x=6\)  is a root:}\)

♦♦♦ Mean mark (ii) 24%.

\(8 x^3-51 x^2+108=(x-6)\left(8 x^2-3 x-18\right)\)

\(\text{Other roots:}\)

  \(x\) \(=\dfrac{3 \pm \sqrt{9-4 \cdot 8 \cdot 18}}{2 \times 8}\)
    \(=\dfrac{3 \pm \sqrt{585}}{16}\)
    \(=\dfrac{3+3 \sqrt{65}}{16} \quad(x>0)\)
    \(=1.7 \ \text{units (1 d.p.)}\)

 
\(\therefore \ \text{Object next comes to rest at  \(x=1.7\) units}\) 

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`
Show Answers Only

`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 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`?

  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 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)`
  `= (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
  3.    `t = (b sqrt (2b))/mu int_0^(cos^-1 sqrt (d/b)) cos^2 theta\ d theta.`   (3 marks)

  4. 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.)
  5. What is the limiting time taken for the particle to reach `O?`   (1 mark)

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a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `(pi(sqrtb)^3)/(2sqrt2mu)\ text(seconds)`

Show Worked Solution

a.    `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.)`

 

b.     `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)`

 

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

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

♦♦ Mean mark (c) 9%.
`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
  3.    `v^2 = (gR^3)/x^2-(gR-u^2).`   (3 marks)

  4. It can be shown that  `x = sqrt (R^2 + 2uRt-(gR-u^2) t^2).` (Do NOT prove this.)
  5. Show that if  `u >= sqrt (gR)`  the particle will not return to the planet.   (2 marks)

  6. If  `u < sqrt (gR)`  the particle reaches a point whose distance from the centre of the planet is `D`, and then falls back.
  7.   i.  Use the formula in part (b) to find `D` in terms of `u, R` and `g.`   (1 mark)

  8.   ii. Use the formula in part (c) 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

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

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

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

d.i   `D = sqrt ((gR^3)/(gR-u^2))`

d.ii   `(2uR)/(gR-u^2)`

Show Worked Solution
a.    

`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`

 

b.     `ddot x` `=- (gR^3)/x^3`
  `1/2 v^2` `= int-(gR^3)/x^3\ dx`
  `1/2 v^2` `= (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)`

 

c.    `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.)`

 

d.i   `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))`

 

d.ii  `text(Find)\ \ t\ \ text(when)\ \ x = R`

`text{Using part (c)}`

`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

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

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

Show Worked Solution

a.    `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 “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)`
 

b.    `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

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)

Show Answers Only

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`

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`
`1/2 v^2` `= 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`