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Calculus, 2ADV C4 2023 HSC 26

A camera films the motion of a swing in a park.

Let \(x(t)\) be the horizontal distance, in metres, from the camera to the seat of the swing at \(t\) seconds.

The seat is released from rest at a horizontal distance of 11.2 m from the camera.
 

  1. The rate of change of \(x\) can be modelled by the equation

\(\dfrac{dx}{dt}=-1.5\pi\ \sin(\dfrac{5\pi}{4}t)\).

  1. Find an expression for \(x(t)\).  (2 marks)

  2. How many times does the swing reach the closest point to the camera during the first 10 seconds?  (2 marks)

Show Answers Only

  1. \(x(t)=\dfrac{6}{5} \cos(\dfrac{5\pi}{4}t) + 10\)
  2. \(\text{6 times}\)

Show Worked Solution

a.    \(\dfrac{dx}{dt}=-1.5\pi\ \sin(\dfrac{5\pi}{4}t)\)

\(x(t)\) \(= -1.5\pi\ \int \sin(\dfrac{5\pi}{4}t)\ dt \)  
  \(= 1.5\pi \times \dfrac{4}{5\pi} \times \cos(\dfrac{5\pi}{4}t) + c\)  
  \(=\dfrac{6}{5} \cos(\dfrac{5\pi}{4}t) + c \)  
Mean mark (a) 51%.

  
\(\text{When}\ t=0, \ x(t)=11.2:\)

\(11.2\) \(=\dfrac{6}{5} \cos(0) + c\)  
\(c\) \(=11.2-\dfrac{6}{5} \)  
  \(=10\)  

 
\(x(t)=\dfrac{6}{5} \cos(\dfrac{5\pi}{4}t) + 10\)

 

b.    \(\text{Period}\ =\dfrac{2\pi}{n} = \dfrac{2\pi}{\frac{5\pi}{4}} = \dfrac{8}{5} = 1.6\ \text{(seconds)}\)
♦♦ Mean mark (b) 31%.

\(\text{1st time swing reaches closest point to camera = 0.8 seconds}\)

\(\text{Periods in next 9.2 seconds}\) \(=\dfrac{9.2}{1.6}\)  
  \(= 5.75\ \text{times}\)  

 
\(\therefore\ \text{Swing reaches the closest point 6 times in the 1st 10 seconds}\)

Calculus, 2ADV C4 EQ-Bank 5

The velocity of a particle moving along the `x`-axis at `v` metres per second at `t` seconds, is shown in the graph below.
 


 

Initially, the displacement `x` is equal to 12 metres.

  1. Write an equation that describes the displacement, `x`, at time `t` seconds.  (2 marks)

  2. Draw a graph that shows the displacement of the particle, `x`  metres from the origin, at a time `t` seconds between  `t= 0`  and  `t = 5`. Label the coordinates of the endpoints of your graph.  (2 marks)

Show Answers Only
  1. `x = 3t – (3)/(10) t ^2 + 12`
  2.  

Show Worked Solution
i.    `m_v` `= -(3)/(5)`
  `v` `= 3 – (3)/(5) t`

 

`x` `= int v \ dt`
  `= int 3 – (3)/(5) t \ dt`
  `= 3t – (3)/(10) t^2 + c`

 
`text(When) \ \ t = 0, x = 12  \ => \ c = 12`

`:. \ x = 3t – (3)/(10) t ^2 + 12`

 

ii.  `text(When) \ \ t = 5:`

`x = 15 – (3)/(10) xx 25 + 12 = 19.5`
 

Calculus, 2ADV C4 SM-Bank 1 MC

A lift accelerates from rest at a constant rate until it reaches a speed of 3 ms−1. It continues at this speed for 10 seconds and then decelerates at a constant rate before coming to rest. The total travel time for the lift is 30 seconds.

The total distance, in metres, travelled by the lift is

  1.  45
  2.  60
  3.  75
  4.  90
Show Answers Only

`B`

Show Worked Solution

`text(Consider the velocity graph:)`
 

`t_1 -> t_2 = text(10 seconds)`
 

`:.\ text(Total distance travelled)`

`=\ text(Area of trapezium)`

`= 1/2 xx 3(10 + 30)`

`= 60\ text(m)`
 

`=>\ B`

Calculus, 2ADV C4 2019 HSC 14a

A particle is moving along a straight line. The particle is initially at rest. The acceleration of the particle at time  `t`  seconds is given by  `a = e^(2t)-4`, where  `t >= 0`.

Find an expression, in terms of  `t`, for the velocity of the particle.  (2 marks)

Show Answers Only

`v = 1/2e^(2t)-4t-1/2`

Show Worked Solution

`a = (dv)/(dt) = e^(2t)-4`

`v` `= int e^(2t)-4\ dt`
  `= 1/2 e^(2t)-4t + c`

 
`text(When)\ t = 0,\ v = 0`

`0 = 1/2 e^0-0 + c`

`c = -1/2`

`:. v = 1/2e^(2t)-4t-1/2`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)

  2. If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)

  3. Sketch the displacement, `x`, as a function of time.  (2 marks)

Show Answers Only
  1. `t > 5\ \ text(seconds)`
  2. `7.2\ \ text(seconds)`
  3.    
Show Worked Solution

i.  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

ii.  `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D:`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 7/5`
  `= 1.4\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.4 seconds.)`
   

iii.   

Calculus, 2ADV C4 2016 HSC 16a

A particle moves in a straight line. Its velocity `v\ text(ms)^-1` at time `t` seconds is given by

`v = 2 - 4/(t + 1).`

  1. Find the initial velocity.  (1 mark)

  2. Find the acceleration of the particle when the particle is stationary.  (2 marks)

  3. By considering the behaviour of `v` for large `t`, sketch a graph of `v` against `t` for  `t >= 0`, showing any intercepts.  (2 marks)

  4. Find the exact distance travelled by the particle in the first 7 seconds.  (3 marks)

Show Answers Only
  1. `-2\ text(ms)^-1`
  2. `1\ text(ms)^-2`
  3.  
    hsc-2016-16ai
  4. `10 – 4 ln 2\ \ text(metres)`
Show Worked Solution

i.   `text(Initial velocity when)\ \ t = 0`

Mean mark 93%.
COMMENT: Great example of low hanging fruit late in the exam for students who progress efficiently.

`v = 2 – 4/1 = -2\ text(ms)^-1`

 

ii.   `v` `= 2 – 4/(t + 1)`
  `a` `=(dv)/(dt)= 4/(t + 1)^2`

 

`text(Particle is stationary when)\ \ v = 0,`

`2 – 4/(t + 1)` `= 0`
`2 (t + 1)` `= 4`
`t` `= 1`
   

`text(When)\ \ t=1,`

`:.a` `= 4/(1 + 1)^2`
  `= 1\ text(ms)^-2` 

 

iii.  `v = 2 – 4/(t + 1)`

♦ Mean mark 47%.

`text(As)\ \ t -> oo,\ \ \ 4/(t + 1) -> 0`

`:. v -> 2`

 hsc-2016-16ai

 

♦♦ Mean mark 26%.

iv.   `text(Distance travelled in 1st 7 seconds)`

`= |\ int_0^1 (2 – 4/(t + 1))\ dt\ | + int_1^7 (2 – 4/(t + 1))\ dt`

`= -[2t – 4 ln (t + 1)]_0^1 + [2t – 4 ln (t + 1)]_1^7`

`= -[(2 – 4 ln 2) – 0] + [(14 – 4 ln 8) – (2 – 4 ln 2)]`

`= 4 ln 2 – 2 + 12 – 4 ln 2^3 + 4 ln 2`

`= 10 + 8 ln 2 – 12 ln 2`

`= 10 – 4 ln 2\ \ text(metres)`

Calculus, 2ADV C4 2015 HSC 9 MC

A particle is moving along the `x`‑axis. The graph shows its velocity `v` metres per second at time `t` seconds.
 

2012 2ua 9 mc
 

When `t = 0` the displacement `x` is equal to `2` metres.

What is the maximum value of the displacement `x`?

  1. `text(8 m)`
  2. `text(14 m)`
  3. `text(16 m)`
  4. `text(18 m)`
Show Answers Only

`D`

Show Worked Solution

`text(Distance travelled)`

♦♦ Mean mark 31%.

`= int_0^4 v\ dt`

`= text(Area under the velocity curve)`

`= 1/2 xx b xx h`

`= 1/2 xx 4 xx 8`

`= 16\ \ text(metres.)`

 

`text(S)text(ince velocity is always positive between)\ t=0`

`text(and)\  t = 4,\ text(and the original displacement = 2,)`

`text(the maximum displacement) = 16 + 2 = 18\ \ text(metres)`

`=> D`

Calculus, 2ADV C4 2009 HSC 7a

The acceleration of a particle is given by

`a=8e^(-2t)+3e^(-t)`,

where  `x`  is the displacement in metres and  `t`  is the time in seconds.

Initially its velocity is  `text(– 6 ms)^(–1)` and its displacement is 5 m.

  1. Show that the displacement of the particle is given by
  2. `qquad  x=2e^(-2t)+3e^-t+t`.   (2 marks) 

  3. Find the time when the particle comes to rest.    (3 marks)

  4. Find the displacement  when the particle comes to rest.    (1 mark)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `ln4\ text(seconds)`
  3. `7/8+ln4\ \ text(units)`
Show Worked Solution

i.    `text(Show)\ \ x=2e^(-2t)+3e^-t+t`

`a=8e^(-2t)+3e^-t\ \ text{(given)}`

`v=int a\ dt=-4e^(-2t)-3e^-t+c_1`

`text(When)\ t=0,  v=-6\ \ text{(given)}`

`-6` `=-4e^0-3e^0+c_1`
`-6` `=-7+c_1`
`c_1` `=1`

 
`:. v=-4e^(-2t)-3e^-t+1`
 

`x` `=int v\ dt`
  `=int(-4e^(-2t)-3e^-t+1)\ dt`
  `=2e^(-2t)+3e^-t+t+c_2`

 
`text(When)\ \ t=0,\ x=5\ \ text{(given)}`

`5` `=2e^0+3e^0+c_2`
`c_2` `=0`

 
`:.\ x=2e^(-2t)+3e^-t+t\ \ text(… as required)`

 

ii.   `text(Particle comes to rest when)\ \ v=0`

`text(i.e.)\ \ -4e^(-2t)-3e^-t+1=0`

`text(Let)\ X=e^-t\ \ \ \ =>X^2=e^(-2t)`

`-4X^2-3X+1` `=0`
`4X^2+3X-1` `=0`
`(4X-1)(X+1)` `=0`

 
 `:.\ \ X=1/4\ \ text(or)\ \ X=-1`

`text(When)\ \ X=1/4:`

`e^-t` `=1/4`
`lne^-t` `=ln(1/4)`
`-t` `=ln(1/4)`
`t` `=-ln(1/4)=ln(1/4)^-1=ln4`

 
`text(When)\ \ X=-1:`

`e^-t=-1\ \ text{(no solution)}`
 

`:.\ text(The particle comes to rest when)\ t=ln4\ text(seconds)`
  

iii.  `text(Find)\ \ x\ \ text(when)\ \ t=ln4 :`

`x=2e^(-2t)+3e^-t+t`

`\ \ =2e^(-2ln4)+3e^-ln4+ln4`

`\ \ =2(e^ln4)^-2+3(e^ln4)^-1+ln4`

`\ \ =2xx4^-2+3xx4^-1+ln4`

`\ \ =2/16+3/4+ln4`

`\ \ =7/8+ln4`

ALGEBRA TIP: Helpful identity  `e^lnx=x`. Easily provable as follows:
`e^ln2=x`
`\ =>lne^ln2=lnx\ `
`\ => ln2=lnx\ `
`\ =>x=2`.

Calculus, 2ADV C4 2012 HSC 15b

The velocity of a particle is given by

`v=1-2cost`,

where  `x`  is the displacement in metres and  `t`  is the time in seconds. Initially the particle is 3 m to the right of the origin.

  1. Find the initial velocity of the particle.    (1 mark)

  2. Find the maximum velocity of the particle.    (1 mark)

  3. Find the displacement, `x`,  of the particle in terms of  `t`.    (2 marks)

  4. Find the position of the particle when it is at rest for the first time.    (2 marks)

Show Answers Only
  1. `-1\ text(m/s)`
  2. `3\ text(m/s)`
  3. `x=t-2sint+3`
  4. `pi/3-sqrt3+3`
Show Worked Solution

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

`v` `=1-2cos0`
  `=1-2`
  `=-1`

 
`:.\ text(Initial velocity is)\ -1\ text(m/s.)`

 

ii.  `text(Solution 1)`

`text(Max velocity occurs when)\ \ a=(d v)/(dt)=0`

♦♦ Mean mark 29%
MARKER’S COMMENT: Solution 2 is more efficient here. Using the -1 and +1 limits of trig functions can be very a effective way to calculate max/min values.

`a=2sint`
 

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

`2sint=0`

`t=0`,  `pi`,  `2pi`, …

`text(At)\ \ t=0,\ \   v=-1\ text(m/s)`

`text(At)\ \ t=pi,\ \ v=1-2(-1)=3\ text(m/s)`

 
`:.\ text(Maximum velocity is 3 m/s)`

 

`text(Solution 2)`

`v=1-2cost`

`text(S)text(ince)\ \ -1` `<cost<1`
`-2` `<2cost<2`
`-1` `<1-2cost<3`

 
`:.\ text(Maximum velocity is 3 m/s)`

 

iii.   `x` `=int v\ dt`
  `=int(1-2cost)\ dt`
  `=t-2sint+c`

 
`text(When)\ \ t=0,\ \ x=3\ \ text{(given)}`

`3=0-2sin0+3`

`c=3`

 
`:. x=t-2sint+3`

 

iv.  `text(Find)\ \ x\ \ text(when)\ \ v=0\ \ text{(first time):}`

♦ Mean mark 50%
MARKER’S COMMENT: Many students found  `t=pi/3`  but failed to gain full marks by omitting to find  `x`. Remember that for calculus, angles are measured in radians, NOT degrees!

`text(When)\ \ v=0 ,`

`0` `=1-2cost`
`cost` `=1/2`
`t` `=cos^-1(1/2)`
  `=pi/3\ \ \ text{(first time)}`

 
`text(Find)\ \ x\ \ text(when)\ \ t=pi/3 :`

`x` `=pi/3-2sin(pi/3)+3`
  `=pi/3-2xxsqrt3/2+3`
  `=pi/3-sqrt3+3\ \ text(units)`