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Calculus, SPEC2 2023 VCAA 13 MC

A tourist in a hot air balloon, which is rising vertically at 2.5 m s\(^{-1}\), accidentally drops a phone over the side when the phone is 80 metres above the ground.

Assuming air resistance is negligible, how long in seconds, correct to two decimal places, does it take for the phone to hit the ground?

  1. 2.86
  2. 2.98
  3. 3.79
  4. 4.04
  5. 4.30
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Take upward velocity as positive.}\)

\(u=2.5\ \text{ms}^{-1}, \ a=-9.8\ \text{ms}^{-2} \)

\(\text{Using}\ \ s=ut+\dfrac{1}{2}at^2,\)

\(\text{Find}\ t\ \text{when}\ \ s=-80\ \text{(by calc):} \)

\(-80=2.5t-4.9t^2\ \ \Rightarrow \ \ t=4.30\ \text{s}\)

\(\Rightarrow E\)

Calculus, SPEC2 2012 VCAA 5

At her favourite fun park, Katherine’s first activity is to slide down a 10 m long straight slide. She starts from rest at the top and accelerates at a constant rate, until she reaches the end of the slide with a velocity of `6\ text(ms)^(-1)`.

  1. How long, in seconds, does it take Katherine to travel down the slide?   (1 mark)

When at the top of the slide, which is 6 m above the ground, Katherine throws a chocolate vertically upwards. The chocolate travels up and then descends past the top of the slide to land on the ground below. Assume that the chocolate is subject only to gravitational acceleration and that air resistance is negligible.

  1. If the initial speed of the chocolate is 10 m/s, how long, correct to the nearest tenth of a second, does it take the chocolate to reach the ground?   (2 marks)

  2. Assume that it takes Katherine four seconds to run from the end of the slide to where the chocolate lands.

     

    At what velocity would the chocolate need to be propelled upwards, if Katherine were to immediately slide down the slide and run to reach the chocolate just as it hits the ground?

     

    Give your answer in `text(ms)^(-1)`, correct to one decimal place.   (2 marks)

Katherine’s next activity is to ride a mini speedboat. To stop at the correct boat dock, she needs to stop the engine and allow the boat to be slowed by air and water resistance.

At time `t` seconds after the engine has been stopped, the acceleration of the boat, `a\ text(ms)^(-2)`, is related to its velocity, `v\ text(ms)^(-1)`, by

`a = -1/10 sqrt(196-v^2)`.

Katherine stops the engine when the speedboat is travelling at `7\ text(m/s)`.

  1. i.  Find an expression for `v` in terms of `t`.   (3 marks)

  2. ii. Find the time it takes the speedboat to come to rest.Give your answer in seconds in terms of `pi`.   (2 marks)

  3. iii. Find the distance it takes the speedboat to come to rest, from when the engine is stopped.Give your answer in metres, correct to one decimal place.   (3 marks)

Show Answers Only
  1. `10/3\ text(s)`
  2. `2.5\ text(s)`
  3. `35.1\ text(m s)^(−1)`
  4. i.  `v = 14 sin(pi/6-t/10)`
  5. ii. `(5 pi)/3`
  6. iii. `18.8`
Show Worked Solution

a.   `u = 0, quad v = 6, quad s = 10`

COMMENT: Exact form required. 3.3 seconds was marked incorrect!

`text(Solve for)\ \ t:\ \ \ ((u + v)/2)t` `= s`
`((0 + 6)/2)t` `= 10`
`t` `= 10/3\ text(s)`

 

b.   `u = 10, quad a = -9.8, quad s = -6`

COMMENT: Many students showed a “lack of understanding” of displacement here.

`text(Solve for)\ \ t:`

`s` `= ut + 1/2 at^2`
`-6` `= 10t-4.9 t^2`
`:. t` `~~2.5\ text(s)\ \ \ text{(by CAS)}`

♦♦ Mean mark 27%.

c.   `text(Time of chocolate in air)`

`= 10/3 + 4`

`=22/3`
 

`text(Solve for)\ \ u:`

`-6` `= 22/3 u -4.9(22/3)^2`
`:. u` `~~35.1\ text(m s)^(−1)\ \ \ text{(by CAS)}`

 

d.i.    `(dv)/(dt)` `= -1/10 sqrt(196-v^2)`
  `(dt)/(dv)` `= (-10)/sqrt(196-v^2)`
  `t` `= int(-10)/sqrt(196-v^2)\ dv`
  `-t/10` `= int 1/sqrt(14^2-v^2) dv`
  `-t/10` `= sin^(-1)(v/14) + c`

 
`text(When)\ \ t = 0, v = 7`

`=> c=-sin^(-1)(1/2) = -pi/6`
 

`-t/10` `=sin^(-1)(v/14)-pi/6`  
`sin^(-1) (v/14)` `= pi/6-t/10`  
`:. v` `=14sin(pi/6-t/10)`  

 

d.ii.  `text(Find)\ \ t\ \ text(when)\ \ v=0:`

  `-t/10` `=sin^(-1)(0)-pi/6`
  `t` `= -10 sin^(-1)(0) + (10 pi)/6`
    `= (5 pi)/3\ text(s)`

 

d.iii.   `v = 14sin(pi/6-t/10)`

`(dx)/(dt)` `=14sin(pi/6-t/10)`
`x` `= int_0^((5pi)/3)14sin(pi/6-t/10)\ dt`
  `~~ 18.8\ text(m)\ \ \ text{(by CAS)}`

Calculus, SPEC2 2013 VCAA 19 MC

A tourist in a hot air balloon, which is rising at 2 m/s, accidentally drops a camera over the side and it falls 100 m to the ground.

Neglecting the effect of air resistance on the camera, the time taken for the camera to hit the ground, correct to the nearest tenth of a second, is

A.   4.3 s

B.   4.5 s

C.   4.7 s

D.   4.9 s

E.   5.0 s

Show Answers Only

`C`

Show Worked Solution

`u = 2, s = −100, a = -9.8`

`s` `= ut + 1/2at^2`
`-100` `= 2t – 4.9t^2`
`0` `= 4.9t^2 – 2t – 100`

 
`text(Solve:)\ \ 4.9t^2 – 2t – 100=0\ \ text(for)\ \ t`

`t~~4.7\ \ text(sec)`

`=> C`

Calculus, SPEC2 2015 VCAA 22 MC

A ball is thrown vertically up with an initial velocity of  `7sqrt6` ms`\ ^(–1)`, and is subject to gravity and air resistance.

The acceleration of the ball is given by  `ddotx = −(9.8 + 0.1v^2)`, where `x` metres is its vertical displacement, and `v` ms`\ ^(–1)` is its velocity at time `t` seconds.

The time taken for the ball to reach its maximum height is

A.   `pi/3`

B.   `(5pi)/(21sqrt2)`

C.   `log_e(4)`

D.   `(10pi)/(21sqrt2)`

E.   `10log_e(4)`

Show Answers Only

`D`

Show Worked Solution

`(dv)/(dt) = −(9.8 + 0.1v^2)`

♦ Mean mark 42%.

`(dt)/(dv) = −1/(9.8 + 0.1v^2)`
 

`text(Max height occurs when)\ \ v=0.`

`text(Find)\ \ t\ \ text(when)\ \ v=0:`

`t` `= int_(7sqrt6)^0 −1/(9.8 + v^2/10)\ dv`
  `= (10pi)/(21sqrt2)`

 
`=> D`