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Mechanics, EXT2 EQ-Bank 30

Luggage at an airport is delivered to its owners via a ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

 

     

  1.  By resolving forces parallel to the ramp, show that the magnitude of the acceleration, `ddot{x}\ text(ms)^(-2)`, of the suitcase down the ramp is given by  `ddot{x} = (g-2v)/2`.   (2 marks)

  2. Using 9.8 `text(ms)^(-2)` as the acceleration due to gravity, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form  `x = bv + c\ log_e(c/(c-v))`, where `b, c in R`.   (3 marks)

Show Answers Only

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

b.    `x = -v + 4.9 ln ((4.9)/(4.9-v))`

Show Worked Solution

a.    
         

`sumF` `=20g sin 30^@-20v`  
`m ddot{x}` `=10g-20v`  
`ddot{x}` `= (g-2v)/2`  

 
b.    `text{Using}\ \ ddot{x}=v *(dv)/(dx):`

`(dv)/(dx)` `= (g-2v)/(2v)`
`(dx)/(dv)` `= (2v)/(g-2v)`
`(dx)/(dv)` `= -(2v)/(2v-g)=-((2v-g + g))/(2v-g)= -1-g/(2v-g)`

 
`text{Find the distance travelled:}`

`x` `= int_0^v-1-g/(2v-g)\ dv`
  `= int_0^v-1-g/2 (2/(2v-g))\ dv`
  `= [-v-4.9 xx ln\ |2v-g|]_0^v`

 
`text(When)\ \ x=0, v=0:`

`2v-g < 0\ \ =>\ \ |2v-g| = g-2v`
 

`x` `= [-v-4.9 ln (g-2v)]_0^v`
  `= -v-4.9 ln (g-2v)-(0-4.9 ln (g))`
  `= -v + 4.9 ln (g/(g-2v))`
  `=-v + 4.9 ln(9.8/(9.8-2v))`
  `= -v + 4.9 ln ((4.9)/(4.9-v))`

Mechanics, EXT2 M1 2025 HSC 12e

A particle of mass \(m\) kg moves along a horizontal line with an initial velocity of \(V_0 \ \text{ms}^{-1}\).

The motion of the particle is resisted by a constant force of \(m k\) newtons and a variable force of \(m v^2\) newtons, where \(k\) is a positive constant and \(v \ \text{ms}^{-1}\) is the velocity of the particle at \(t\) seconds.

Show that the distance travelled when the particle is brought to rest is  \(\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\) metres.   (3 marks)

Show Answers Only
\(ma\) \(-m k-m v^2\)
\(a\) \(=-k-v^2\)
\(v \cdot \dfrac{d v}{d x}\) \(=-\left(k+\nu^2\right)\)
\(\dfrac{d x}{d v}\) \(=-\dfrac{v}{k+v^2}\)
\(x\) \(=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c\)

  
\(\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)\)

\(x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)\)

\(\text{Find \(x\) when \(\ v=0\ \) (distance travelled):}\)

\(x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}\)

Show Worked Solution
\(ma\) \(-m k-m v^2\)
\(a\) \(=-k-v^2\)
\(v \cdot \dfrac{d v}{d x}\) \(=-\left(k+\nu^2\right)\)
\(\dfrac{d x}{d v}\) \(=-\dfrac{v}{k+v^2}\)
\(x\) \(=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c\)

 
\(\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)\)

\(x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)\)
 

\(\text{Find \(x\) when \(\ v=0\ \) (distance travelled):}\)

\(x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}\)

Mechanics, EXT2 M1 2024 HSC 13c

A particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to \(v^2\), where \(v\) m s\(^{-1}\) is the speed of the particle, so that the acceleration is given by  \(-k v^2\).

Initially the particle is at the origin and has a velocity of 40 m s\(^{-1}\) to the right. After the particle has moved 15 m to the right, its velocity is 10 m s\(^{-1}\) (to the right).

  1. Show that  \(v=40 e^{-k x}\).   (3 marks)

  2. Show that  \(k=\dfrac{\ln 4}{15}\).   (1 mark)

  3. At what time will the particle's velocity be 30 m s\(^{-1}\) to the right?   (3 marks)

Show Answers Only

i.    \(\ddot{x}=v \cdot \dfrac{d v}{d x}=-k v^2\)

  \(\dfrac{d v}{d x}\) \(=-k v\)
  \(\dfrac{d x}{d v}\) \(=-\dfrac{1}{k v}\)
  \(\displaystyle\int \frac{1}{v}\, d v\) \(=\displaystyle -\int k\, d x\)
  \(\ln \abs{v}\) \(=-k x+c\)
  \(v\) \(=e^{-k x+c}\)
    \(=A e^{-k x}\ \ (\text{where } A=e^c)\)

 
\(\text {When } x=0, v=40:\)

\(40=A e^{\circ} \ \Rightarrow \ A=40\)

\(\therefore V=40 \, e^{-k x}\)
 

ii.   \(\text {Show}\ \ k=\dfrac{\ln 4}{15}\)

\(\text {When } x=15, v=10:\)

  \(10\) \(=40 e^{-15 k}\)
  \(e^{-15 k}\) \(=\dfrac{1}{4}\)
  \(-15 k\) \(=\ln \left(\dfrac{1}{4}\right)\)
  \(15 k\) \(=\ln 4\)
  \(k\) \(=\dfrac{\ln 4}{15}\)

iii.  \(\dfrac{1}{8\,\ln 4}\  \text{seconds}\)

Show Worked Solution

i.    \(\ddot{x}=v \cdot \dfrac{d v}{d x}=-k v^2\)

  \(\dfrac{d v}{d x}\) \(=-k v\)
  \(\dfrac{d x}{d v}\) \(=-\dfrac{1}{k v}\)
  \(\displaystyle\int \frac{1}{v}\, d v\) \(=\displaystyle -\int k\, d x\)
  \(\ln \abs{v}\) \(=-k x+c\)
  \(v\) \(=e^{-k x+c}\)
    \(=A e^{-k x}\ \ (\text{where } A=e^c)\)

 
\(\text {When } x=0, v=40:\)

\(40=A e^{\circ} \ \Rightarrow \ A=40\)

\(\therefore V=40 \, e^{-k x}\)
 

ii.   \(\text {Show}\ \ k=\dfrac{\ln 4}{15}\)

\(\text {When } x=15, v=10:\)

  \(10\) \(=40 e^{-15 k}\)
  \(e^{-15 k}\) \(=\dfrac{1}{4}\)
  \(-15 k\) \(=\ln \left(\dfrac{1}{4}\right)\)
  \(15 k\) \(=\ln 4\)
  \(k\) \(=\dfrac{\ln 4}{15}\)

 

iii.   \(\text {Find}\ t\ \text {when}\ \ v=30:\)

  \(\dfrac{d v}{d t}\) \(=-k v^2\)
  \(\dfrac{d t}{d v}\) \(=-\dfrac{1}{k v^2}\)
  \(t\) \(=\displaystyle -\int \dfrac{1}{k v^2}\, d v=\dfrac{1}{k v}+c\)

 
\(\text {When}\ \ t=0, v=40:\)

\(0=\dfrac{1}{40 k}+c \ \Rightarrow \ c=-\dfrac{1}{40 k}\)
 

\(\text{Find \(t\) when  \(v=30\):}\)

  \(t\) \(=\dfrac{1}{30k}-\dfrac{1}{40k}\)
    \(=\dfrac{1}{120k}\)
    \(=\dfrac{15}{120\, \ln 4}\)
    \(=\dfrac{1}{8\,\ln 4}\  \text{seconds}\)

Mechanics, EXT2 M1 2022 HSC 12c

A particle with mass 1 kg is moving along the `x`-axis. Initially, the particle is at the origin and has speed `u` m s-1 to the right. The particle experiences a resistive force of magnitude  `v+3 v^2`  newtons, where `v` m s-1 is the speed of the particle after `t` seconds. The particle is never at rest.

  1. Show that  `(dv)/(dx)=-(1+3v)`   (1 mark)

  2. Hence, or otherwise, find `x` as a function of `v`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `x=1/3 ln((1+3u)/(1+3v))`
Show Worked Solution

i.    `F=m ddotx=ddotx\ \ (m=1)`

`ddotx` `=-(v+3v^2)`  
`v*(dv)/(dx)` `=-(v+3v^2)`  
`:. (dv)/(dx)` `=-(1+3v)\ \ text{… as required}`  

  

ii.     `(dx)/(dv)` `=- 1/(1+3v)`
  `x` `=-int1/(1+3v)\ dv`
    `=-1/3 ln(1+3v)+c`

 
`text{When}\ \ t=0, \ v=u\ \ =>\ \ c=1/3 ln(1+3u)`

`:.x` `=1/3 ln(1+3u)-1/3 ln(1+3v)`  
  `=1/3 ln((1+3u)/(1+3v))`  

Mechanics, EXT2 M1 2020 HSC 14b

A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2, is modelled by

`a = 10 (1-(kv)^2)`,

where `v` is the velocity of the particle in m/s and  `k = 0.01`.

Find the velocity of the particle after 5 seconds.   (4 marks)

Show Answers Only

`46.21 \ text{ms}^-1`

Show Worked Solution
`a` `= 10 (1-(kv)^2)`
`frac{dv}{dt}` `= 10-10 xx 0.01^2 xx v^2= 10-0.001 v^2`
`frac{dt}{dv}` `= frac{1}{10-0.001 \ v^2}= frac{1000}{10 \ 000-v^2}`
`t` `= int frac{1000}{100^2-v^2}\ dv`

 
`text{Using partial fractions}:`

`frac{1}{100^2-v^2}` ` = frac{A}{100 + v} + frac{B}{100-v}`
`1` `= A (100-v) + B(100 + v)`

 
`text{If} \ \ v = 100 \ , \ 1 = 200 B \ => \ B = frac{1}{200}`

`text{If} \ \ v = -100 \ , \ 1 = 200 A \ => \ A = frac{1}{200}`
 

`t` `= 1000 int frac{1}{200 (100 + v)}\ dv + 1000 int frac{1}{200(100 -v)}\ dv`
  `= 5 int frac{1}{100 + v}\ dv +  5 int frac{1}{100-v}\ dv`
  `= 5 ln \ | 100 + v |-5 ln  \|100-v | + c`
  `= 5 ln \ | frac{100 + v}{100-v} | + c`

 
`text{When} \ \ t = 0 , \ v = 0`

`0 = 5 ln 1  + c  \ => \ c = 0`
 

`:. t = 5 ln \ | frac{100 + v}{100-v} |`
  

`text{Find} \ v \ \ text{when} \ t = 5 :`

`5` `= 5 ln | frac{100 + v}{100-v} |`
`1` `= ln | frac{100 + v}{100-v} | `
`e` `= frac{100 + v}{100-v}`
`100 e-ve` `= 100 + v`
`v + ve` `= 100 e-100`
`v(1 + e)` `= 100 e-100`
`:. v` `= frac{100 e-100}{1 + e}= 46.21 \ text{ms}^-1 \ \ (text{2 d.p.})`

Mechanics, EXT2 M1 EQ-Bank 26

A torpedo with a mass of 80 kilograms has a propeller system that delivers a force of `F` on the torpedo, at maximum power. The water exerts a resistance on the torpedo proportional to the square of the torpedo's velocity `v`.

  1. Explain why  `(dv)/(dt) = 1/80 (F-kv^2)`
  2. where `k` is a positive constant.   (1 mark)
  3. If the torpedo increases its velocity from `10\ text(m s)^{-1}` to `20\ text(m s)^{-1}`, show that the distance it travels in this time, `d`, is given by
  4. `d = 40/k log_e((F-100k)/(F-400k))`   (3 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

Show Worked Solution

a.    `R ∝ v^2\ \ =>\ \ R = -kv^2\ \ (k\ text{is positive constant})`

`text(Newton’s 2nd Law:)`

`text(Net Force)\ = mddotx = F-R`

`80ddotx` `= F-kv^2`
`ddotx` `= 1/80 (F-kv^2)`
`(dv)/(dt)` `= 1/80 (F-kv^2)`

 

b.     `v · (dv)/(dx)` `= 1/80 (F-kv^2)`
  `(dv)/(dx)` `= (F-kv^2)/(80v)`
  `(dx)/(dv)` `= (80v)/(F-kv^2)`
  `x` `= −40 int (−2v)/(F-kv^2)\ dv`
    `= −40/k log_e (F-kv^2) + C`

 
`text(When)\ \ v = 10:`

`x_1 = −40/k log_e (F-100k) + C`
  

`text(When)\ \ v = 20:`

`x_2 = −40/k log_e (F-400k) + C`
 

`d` `= x_2-x_1`
  `= −40/k log_e (F-400k) + 40/k log_e (F-100k)`
  `= 40/k log_e ((F-100k)/(F-400k))`

Mechanics, EXT2 M1 EQ-Bank 36

A particle with mass `m` moves horizontally against a resistance force `F`, equal to  `mv(1 + v^2)`  where `v` is the particle's velocity.

Initially, the particle is travelling in a positive direction from the origin at velocity `T`.

  1. Show that the particle's displacement from the origin, `x`, can be expressed as
  2.     `x = tan^(-1)((T-v)/(1 + Tv))`   (2 marks)

  3. Show that the time, `t`, when the particle is travelling at velocity `v`, is given by
  4.     `t = 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`   (4 marks)

  5. Express `v^2` as a function of `t`, and hence find the limiting values of `x` and `v`.   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `v -> 0, \ x -> tan^(−1)T`

Show Worked Solution

a.    `text(Newton’s 2nd law:)`

`F = mddotx = −mv(1 + v^2)`

`v · (dv)/(dx)` `= −v(1 + v^2)`
`(dv)/(dx)` `= −(1 + v^2)`
`(dx)/(dv)` `= −1/(1 + v^2)`
`x` `= −int 1/(1 + v^2)\ dv= −tan^(−1) v + C`

 
`text(When)\ \ x = 0, v = T:`

`C = tan^(−1)T`

`x = tan^(−1)T-tan^(−1)v`
 

`text(Let)\ \ x = A-B:`

`A = tan^(−1) T \ => \ T = tan A`

`B = tan^(−1)v \ => \ v = tan B`

`tan x= tan(A-B)v= (tan A-tan B)/(1 + tan A tan B)= (T-v)/(1 + Tv)`

`:. x= tan^(−1)((T-v)/(1 + Tv))`
 

b.    `text(Using)\ \ ddotx = (dv)/(dt):`

`(dv)/(dt) = −1/(v(1 + v^2))\ \ =>\ \ t = −int 1/(v(1 + v^2)) dv`
 

`text(Using Partial Fractions):`

`1/(v(1 + v^2)) = A/v + (Bv + C)/(1 + v^2)`

`A(1 + v^2) + (Bv + C)v = 1`

`A = 1`

`(A + B)v^2` `= 0`   `=> `    `B` `= −1`
`Cv` `= 0`   `=> `    `C` `= 0`

 

`t` `= −int 1/v\ dv + int v/(1 + v^2)\ dv`
  `= −log_e v + 1/2 int(2v)/(1 + v^2)\ dv`
  `= −log_e v + 1/2 log_e (1 + v^2) + C`
  `= −1/2 log_e v^2 + 1/2 log_e (1 + v^2) + C`
  `= 1/2 log_e ((1 + v^2)/(v^2)) + C`

 

`text(When)\ \ t = 0, v = T:`

`C = −1/2 log_e ((1 + T^2)/(T^2))`
 

`:. t` `= 1/2 log_e ((1 + v^2)/(v^2))-1/2 log_e((1 + T^2)/(T^2))`
  `= 1/2 log_e (((1 + v^2)/(v^2))/((1 + T^2)/(T^2)))`
  `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`

 

c.     `t` `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`
  `e^(2t)` `= (T^2(1 + v^2))/(v^2(1 + T^2))`
  `1 + v^2` `= (e^(2t)v^2(1 + T^2))/(T^2)`
  `1` `= v^2((e^(2t)(1 + T^2))/(T^2)-1)`
  `1` `= v^2((e^(2t)(1 + T^2)-T^2)/(T^2))`
  `:. v^2` `= (T^2)/(e^(2t)(1 + T^2)-T^2)`

 
`text(As)\ \ t -> ∞: \ v^2 -> 0, \ v -> 0`

`x = tan^(−1)T-tan^(−1)v: \ x -> tan^(−1)T`

Mechanics, EXT2 M1 2015 HSC 15a

A particle `A` of unit mass travels horizontally through a viscous medium. When  `t = 0`, the particle is at point `O` with initial speed `u`. The resistance on particle `A` due to the medium is `kv^2`, where `v` is the velocity of the particle at time `t` and `k` is a positive constant.

When  `t = 0`, a second particle `B` of equal mass is projected vertically upwards from `O` with the same initial speed `u` through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle `B` is `kw^2`, where `w` is the velocity of the particle `B` at time `t`. The acceleration due to gravity is `g`.

  1. Show that the velocity `v` of particle `A` is given by  
  2.     `1/v = kt + 1/u.`   (2 marks)

  3. By considering the velocity  `w`  of particle  `B`, show that
  4.     `t = 1/sqrt(gk) (tan^-1(u sqrt(k/g))-tan^-1 (w sqrt(k/g))).`   (3 marks)

  5. Show that the velocity  `V`  of particle  `A`  when particle  `B`  is at rest is given by
  6.     `1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).`   (1 mark)

  7. Hence, if  `u`  is very large, explain why  
  8.     `V ~~ 2/pi sqrt(g/k).`   (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.    `text(See Worked Solutions)`

Show Worked Solution

a.    `text(Particle)\ A:`

`ddot x` `= -kv^2`
`(dv)/(dt)` `= -kv^2`
`(dt)/(dv)` `=- 1/(kv^2)`
`t` `= -1/k int 1/v^2\ dv`
`-kt` `= -1/v + c`

 
`text(When)\ \ t=0,\ \ v=u\ \ => \ \ c=1/u`

`-kt` `= -1/v + 1/u`
`:.1/v` `= kt + 1/u`

 
b.    `text(Particle)\ B:`

`ddot x` `= -g-kw^2`
`(dw)/(dt)` `= -g-kw^2`
`(dt)/(dw)`   `=-1/(g + kw^2)`
`t`   `=-int (dw)/(g + kw^2)`
   `= -1/k int (dw)/(g/k + w^2)`
   `= -1/k xx 1/sqrt (g/k) tan^-1(w/sqrt (g/k)) + c`
  `= -1/sqrt (gk)\ tan^-1 ((sqrt k w)/sqrt g) + c`

 

`text(When)\ \ t = 0,\ w = u:`

`=>c= 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`

`:. t` `= -1/sqrt (gk) tan^-1 ((sqrt k w)/sqrt g) + 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`
  `=1/sqrt (gk) (tan^-1 (u sqrt(k/g))-1/sqrt (gk) tan^-1 (w sqrt (k/g)))`

 

c.    `B\ \ text(at rest when)\ \ w = 0`

`t = 1/sqrt (gk) (tan^-1 (u sqrt (k/g)))`

`:.1/V` `= k xx 1/sqrt(gk) tan^-1 (u sqrt (k/g)) + 1/u,\ \ \ \ text{(part (a))}`
  `= 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g))`

 

d.    `1/V = 1/u + sqrt (k/g) tan^-1 (u sqrt (k/g))`

`text(As)\ \ u -> oo,\ \ tan^-1 (u sqrt (k/g)) -> pi/2`

`:.\ text(If)\ u\ text(is very large,)`

`1/V` `~~ 0 + sqrt (k/g) xx pi/2`
`:.V` `~~ 2/pi sqrt (g/k)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time  `t = 0`  and immediately sinks. Let `x` be its displacement in metres in a downward direction from the surface at time `t` seconds.

The equation of motion is given by

`(dv)/(dt) = 10-(v^2)/40`,

where `v` is the velocity of the object.  

  1. Show that  `v = (20(e^t-1))/(e^t + 1)`.   (4 marks)

  2. Use  `(dv)/(dt) = v (dv)/(dx)`  to show that  
  3.     `x = 20\ log_e(400/(400-v^2))`   (2 marks)

  4. How far does the object sink in the first 4 seconds?   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions.)`

b.    `text(See Worked Solutions.)`

c.    `40log_e((e^4 + 1)/(2e^2))\ text(m)`

Show Worked Solution
a.     `(dv)/(dt)` `= 10-(v^2)/40`
  `(dv)/(dt)` `= (400-v^2)/40`
  `dt` `= 40/(400-v^2)\ dv`
  `int dt` `=int 40/(400-v^2)\ dv`
  `t` `= int (1/(20 + v) + 1/(20-v))\ dv`
    `= log_e(20 + v)-log_e(20-v) + c`

 

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=` ` log_e((20 + v)/(20-v))`
`e^t=` ` (20 + v)/(20-v)`
`20e^t-ve^t=` ` 20 + v`
`v+ ve^t=` ` 20e^t-20`
`v(1+e^t)=` `20(e^t-1)`
`v=` ` (20(e^t-1))/(e^t + 1)\ \ \ \ text(… as required)`

 

b.       `v (dv)/(dx)` `= 10-(v^2)/40`
  `(40v\ dv)/(400-v^2)` `= dx`
`int dx` `= int (40v)/(400-v^2)\ dv`
 `x` `= -20log_e(400-v^2) + c`

 

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x` `= 20log_e400-20log_e(400-v^2)`
  `= 20log_e((400)/(400-v^2))\ \ \ \ text(… as required)`

 

c.    `text(When)\ \ t = 4,\  v = (20(e^4-1))/(e^4 + 1)`

`x` `= 20log_e[400/(400-((20(e^4-1))/(e^4 + 1))^2)]`
  `= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2-(e^4-1)^2)]`
  `= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4-1)(e^4 + 1-e^4 + 1)))`
  `= 20log_e(((e^4 + 1)^2)/(4e^4))`
  `= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Mechanics, EXT2 M1 2014 HSC 14c

A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.

The train is driven by a constant force `F` in the forward direction. The resistive force in the opposite direction is `Kv^2`, where `K` is a positive constant. The terminal velocity of the train is 300 km/h.

  1. Show that the equation of motion for the train is
  2.     `m ddot x = F[1-(v/300)^2]`.   (2 marks)

  3. Find, in terms of `F` and `m`, the time it takes the train to reach a velocity of 200 km/h.   (4 marks)

Show Answers Only

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

b.    `(150 m ln5)/F`

Show Worked Solution

a.    `m ddot x = F-Kv^2,\ \ v_T = 300`

`text(At)\ \ v_T,\ \ ddotx=0:`

`m(0)=F-K(300^2)\ \ =>\ \ K=F/300^2`

`:. m ddot x= F-F/300^2 v^2=F[1-(v/300)^2]\ \ \ text(… as required)`
 

b.     `m*(dv)/(dt)` `= F[1-(v/(300))^2]`
  `(dv)/(1-(v/300)^2)` `=F/m\ dt`
  `(dv)/(300^2-v^2)` `=F/(300^2 m)\ dt`
  `dt` `=(300^2 m)/F xx (dv)/(300^2-v^2)`

 

`int_0^t dt` `=(300 m)/F  int_0^200  300/((300+v)(300-v))\ dv` 
 `:. t` `=(300 m)/F  int_0^200  (1/2)/(300+v) + (1/2)/(300-v)\ dv`
  `=(150 m)/F [ln(300+v)-ln(300-v)]_0^200`
  `=(150 m)/F [ln ((300+v)/(300-v))]_0^200`
  `=(150 m)/F  (ln5-ln1)`
  `=(150 m ln5)/F`