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Mechanics, EXT2 M1 2019 HSC 14b

A parachutist jumps from a plane, falls freely for a short time and then opens the parachute. Let t be the time in seconds after the parachute opens, `x(t)`  be the distance in metres travelled after the parachute opens, and  `v(t)`  be the velocity of the parachutist in `text(ms)^(-1)`.

The acceleration of the parachutist after the parachute opens is given by

`ddot x = g - kv,`

where `g\ text(ms)^(-2)` is the acceleration due to gravity and `k` is a positive constant.

  1. With an open parachute the parachutist has a terminal velocity of  `w\ text(ms)^(-1)`.

     

    Show that  `w = g/k`.  (1 mark)

    At the time the parachute opens, the speed of descent is `1.6 w\ text(ms)^(-1)`.

  2. Show that it takes  `1/k log_e 6`  seconds to slow down to a speed of `1.1w\ text(ms)^(-1)`.  (4 marks)

  3. Let  `D`  be the distance the parachutist travels between opening the parachute and reaching the speed `1.1w\ text(ms)^(-1)`.

     

     

    Show that  `D = g/k^2 (1/2 + log_e 6)`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.   `v_T=w\ \  text(when)\ \ ddot x = 0`

`0` `= g – kw`
`w` `= g/k`

 

ii.   `text(Show)\ \ t = 1/k log_e 6\ \ text(when)\ \ v = 1.1w`

`(dv)/(dt)` `= g – kv`
`(dt)/(dv)` `= 1/(g – kv)`
`t` `= int 1/(g – kv)\ dv`
  `= -1/k ln(g – kv) + C`

 
`text(When)\ \ t = 0,\ \ v = 1.6w`

`0` `= -1/k ln(g – 1.6 kw) + C`
`C` `= 1/k ln(g – 1.6 kw)`
`t` `= 1/k ln (g – 1.6kw) – 1/k ln(g – kv)`
  `= 1/k ln((g – 1.6 kw)/(g – kv))`

 
`text(Find)\ \ t\ \ text(when)\ \ v = 1.1w`

`t` `= 1/k ln((g – 1.6 k xx g/k)/(g – 1.1k xx g/k))`
  `=1/k ln((g – 1.6 g)/(g – 1.1g))`
  `=1/k((-0.6g)/(-0.1g))`
  `= 1/k ln 6`

 

iii.    `v ⋅ (dv)/(dx)` `= g – kv`
  `(dv)/(dx)` `= (g – kv)/v`
  `(dx)/(dv)` `= v/(g – kv)`
  `x` `= int v/(g – kv)\ dv`
    `= 1/k int (kv)/(g – kv)\ dv`
    `= -1/k int 1 – g/(g – kv)\ dv`

 

`:. D` `= -1/k int_(1.6w)^(1.1w) 1 – g/(g – kv)\ dv`
  `= 1/k int_(1.1w)^(1.6w) 1 – g/(g – kv)\ dv`
  `= 1/k[v + g/k ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[(kv)/g + ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[((1.6kw)/g + ln (g – 1.6kw)) – ((1.1 kw)/g + ln (g – 1.1kw))]`
  `= g/k^2[1.6 + ln ((g – 1.6kw)/(g – 1.1kw)) – 1.1]`
  `= g/k^2(0.5 + ln 6)`

Mechanics, EXT2 M1 2011 HSC 6a

Jac jumps out of an aeroplane and falls vertically. His velocity at time  `t`  after his parachute is opened is given by  `v(t)`, where  `v(0) = v_0`  and  `v(t)`  is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is  `kv^2`, where  `k`  is a positive constant. Let  `m`  be Jac’s mass and  `g`  the acceleration due to gravity. Jac’s terminal velocity with the parachute open is  `v_T.`

Jac’s equation of motion with the parachute open is

`m (dv)/(dt) = mg - kv^2.`   (Do NOT prove this.)

  1. Explain why Jac’s terminal velocity  `v_T`  is given by  
     
         `sqrt ((mg)/k).`  (1 mark)

  2. By integrating the equation of motion, show that  `t`  and  `v`  are related by the equation
     
         `t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].`  (3 marks)

  3. Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.

     

    Jac opens his parachute when his speed is  `1/3 v_T.` Gil opens her parachute when her speed is  `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards  `v_T.`

     

    Show that in the time taken for Jac's speed to double, Gil's speed has halved.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `m (dv)/(dt) = mg – kv^2`

`text(As)\ \ v ->text(terminal velocity,)\ \ (dv)/(dt) -> 0`

`mg – kv_T^2` `= 0`
`v_T^2` `= (mg)/k`
`v_T` `= sqrt ((mg)/k)`

  

ii.  `m (dv)/(dt) = mg – kv^2`

`int_0^t dt` `=int_(v_0)^v m/(mg – kv^2)\ dv`
`t` `= m/k int_(v_0)^v (dv)/((mg)/k – v^2)\ \ \ \ text{(using}\ \ v_T^2= (mg)/k text{)}`
  `= (v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`

  

♦ Mean mark part (ii) 39%.

`text(If)\ \ v_0 < v_T,\ \ text(then)\ \ v < v_T\ \ text(at all times.)`

`:. t` `=(v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`
  `=(v_T^2)/g int_(v_0)^v (dv)/((v_T – v)(v_T + v))`
  `=(v_T^2)/(2 g v_T) int_(v_0)^v (1/((v_T – v)) + 1/((v_T + v))) dv`
  `=v_T/(2g)[-ln (v_T – v) + ln (v_T + v)]_(v_0)^v`
  `=v_T/(2g)[ln(v_T + v)-ln(v_T – v)-ln(v_T + v_0)+ln(v_T – v_0)]`
  `=v_T/(2g) ln[((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

 

`text(If)\ \ v_0 > v_T,\ \ text(then)\ \ v > v_T\ \ text(at all times.)`

`text(Replace)\ \ v_T – v\ \ text(by) – (v – v_T)\ \ text(in the preceeding)`

`text(calculation, leading to the same result.)`

 

iii.  `text{From (ii) we have}:\ \ t = v_T/(2g) ln [((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(For Jac,)\ \ v_0 = V_T/3\ \ text(and we have to find the time)`

`text(taken for his speed to be)\ \ v = (2v_T)/3.`

`t` `=v_T/(2g) ln[((v_T + (2v_T)/3)(v_T – v_T/3))/((v_T – (2v_T)/3)(v_T + v_T/3))]`
  `=v_T/(2g) ln [(((5v_T)/3)((2v_T)/3))/((v_T/3)((4v_T)/3))]`
  `=v_T/(2g) ln[(10/9)/(4/9)]`
  `=v_T/(2g) ln (5/2)`

 

`text(For Gil),\ \ v_0 = 3v_T\ \ text(and we have to find)` 

♦♦♦ Mean mark part (iii) 16%.

`text(the time taken for her speed to be)\ \ v = (3v_T)/2.`

`t` `=v_T/(2g) ln [((v_T + (3v_T)/2)(v_T – 3v_T))/((v_T – (3v_T)/2)(v_T + 3v_T))]`
  `=v_T/(2g) ln [(((5v_T)/2)(-2v_T))/((-v_T/2)(4v_T))]`
  `=v_T/(2g) ln [(-5)/-2]`
  `=v_T/(2g) ln (5/2)`

 

`:.\ text(The time taken for Jac’s speed to double is)`

`text(the same as it takes for Gil’s speed to halve.)`