SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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 C3 2021 HSC 26

A particle is shot vertically upwards from a point 100 metres above ground level.

The position of the particle, `y` metres above the ground after `t` seconds, is given by

`y(t) = −5t^2 + 70t + 100`.

  1. Find the maximum height above ground level reached by the particle.  (2 marks)

  2. Find the velocity of the particle, in metres per second, immediately before it hits the ground, leaving your answer in the form  `asqrtb`,  where  `a`  and  `b`  are integers.  (3 marks)

Show Answers Only
  1. `345\ text(m)`
  2. `-20sqrt69\ text(m/s)`
Show Worked Solution

a.    `y(t) = -5t^2 + 70t + 100`

`y^(′)(t) = -10t + 70`

`y^(″)(t) = -10`

`text(Max height occurs when)\ \ y^(′)(t) = 0:`

`-10t + 70` `= 0`
`10t` `= 70`
`t` `= 7`

 

`:.\ text(Max height)\ ` `= -5(7)^2 + 70 xx 7 + 100`
  `= 345\ text(m)`

 

b.   `text(Particle hits ground when)\ \ y = 0:`

♦ Mean mark 38%.
`0` `= -5t^2 + 70t + 100`
`0` `= t^2 – 14t – 20`

 
`text(Using quadratic formula:)`

`t` `= (14 ± sqrt(14^2 + 4*20))/2`
  `= (14 + sqrt(276))/2\ \ \ (t > 0)`
  `= 7 + sqrt69`

 

`:. text(Velocity)\ (y^(′)(t))` `= -10(7 + sqrt69) + 70`
  `= -10sqrt69\ text(m/s)`

 
`:.a=-10 and b=69`

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

a.    `x = 3t – (3)/(10) t ^2 + 12`

b.

   

Show Worked Solution
a.     `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`

 

b.    `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

a.    `t > 5\ \ text(seconds)`

b.    `7.2\ \ text(seconds)`

c.    
       

Show Worked Solution

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

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

 

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

`text(Consider the displacement from)\ B\ text(to)\ D:`

`text{Since the area below the graph from}\ B\ \text(to)\ C\ \text{equals the area}`

`\text{above the graph from}\ C\ \text(to)\ D, \text(there is no change in)`

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

 

`text(Consider)\ 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.)`
   

c.   

     

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 C3 2004 HSC 5b

A particle moves along a straight line so that its displacement, `x` metres, from a fixed point `O` is given by  `x = 1 + 3 cos 2t`, where  `t`  is measured in seconds.

  1. What is the initial displacement of the particle?  (1 mark)

  2. Sketch the graph of  `x`  as a function of  `t`  for  `0 ≤ t ≤ pi`.  (2 marks)

  3. Hence, or otherwise, find when AND where the particle first comes to rest after  `t = 0`.  (2 marks)

  4. Find a time when the particle reaches its greatest magnitude of velocity. What is this velocity?  (2 marks)

Show Answers Only

a.    `text(4 m to the right of)\ O.`

b.    `text(See Worked Solutions)`

c.     `t = pi/2\ text(seconds, 2 m to the left of)\ O.`

d.    `text(6 m s)^(−1)`

Show Worked Solution

a.    `x = 1 + 3 cos 2t`

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

`x` `= 1 + 3 cos 0`
  `= 1 + 3`
  `= 4`

 

`:.\ text(Initial displacement is 4 m to the right of)\ O.`

 

b.    `text(Period)\ = (2pi)/n = (2pi)/2 = pi`

`text(Considering the range)`

`-1` `<=cos 2t<=1`
`-3` `<=3cos 2t<=3`
`-2` `<=1 + 3 cos 2t<=4`

 

 Calculus in the Physical World, 2UA 2004 HSC 5b

 

c.      `x` `= 1 + 3 cos 2t`
  `:.v` `= −6 sin 2t`

 

`text(The particle comes to rest when)\ \ v=0`

`-6 sin 2t` `= 0`
`sin 2t` `= 0`
`2t` `= 0, pi, 2pi…`
`t` `= 0, pi/2, pi…`

 

`:.\ text(After)\ \ t=0, text(particle first comes to rest when)`

`t = pi/2\ text(seconds.)`

`text(When)\ t = pi/2,`

`x` `= 1 + 3 cos 2(pi/2)`
  `= 1 + 3 cos pi`
  `= 1 + 3(−1)`
  `= −2`

 

`:.\ text(Particle first comes to rest at 2 m to the left of)\ O.`

 

d.    `x = 1 + 3 cos 2t`

`v` `= -6 sin 2t`
`a` `= -12 cos 2t`
   

`text(MAX occurs when)\ \ a=0`

`−12 cos 2t` `= 0`
`cos 2t` `= 0`
`2t` `= pi/2, (3pi)/2, …`
`t` `= pi/4, (3pi)/4, …`

 
`:.\ text(Maximum at)\ \ t=pi/4,\ \ (3pi)/4, …\ text(seconds,)`

 

`text(When)\ \ t = pi/4\ text(seconds,)`

`v` `= -6 sin 2(pi/4)`
  `= -6 sin(pi/2)`
  `= −6`

 
`:.\ text(Maximum is 6 m s)^(−1).`

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

a.    `-1\ text(m/s)`

b.    `3\ text(m/s)`

c.    `x=t-2sint+3`

d.    `pi/3-sqrt3+3`

Show Worked Solution

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

`v` `=1-2cos0`
  `=1-2`
  `=-1`

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

 

b.   `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)`

 

c.    `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`

 

d.    `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)`

Calculus, 2ADV C3 2011 HSC 7b

The velocity of a particle moving along the `x`-axis is given by

`v=8-8e^(-2t)`,

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

  1. Show that the particle is initially at rest.     (1 mark)

  2. Show that the acceleration of the particle is always positive.     (1 mark)

  3. Explain why the particle is moving in the positive direction for all  `t>0`.     (2 marks)

  4. As  `t->oo`, the velocity of the particle approaches a constant.

     

    Find the value of this constant.     (1 mark) 

  5. Sketch the graph of the particle's velocity as a function of time.     (2 marks)

Show Answers Only

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

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

c.    `text(See Worked Solutions.)`

d.    `8\ text(m/s)`

e.    `text(See sketch in Worked Solutions)`

Show Worked Solution

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

`v` `=8-8e^0`
  `=0\ text(m/s)`
 
`:.\ text(Particle is initially at rest.)`

 

 

MARKER’S COMMENT: Students whose working showed `e^(-2t)` as `1/e^(2t)`, tended to score highly in this question.

b.    `a=d/(dt) (v)=-2xx-8e^(-2t)=16e^(-2t)`

`text(S)text(ince)\  e^(-2t)=1/e^(2t)>0\ text(for all)\  t`.

`=>\ a=16e^(-2t)=16/e^(2t)>0\ text(for all)\  t`.
 

`:.\ text(Acceleration is positive for all)\ \ t>0`.
 

c.    `text{S}text{ince the particle is initially at rest, and ALWAYS}`

♦♦♦ Mean mark 22%
COMMENT: Students found part (iii) the most challenging part of this question by far.

`text{has a positive acceleration.`
 

`:.\ text(It moves in a positive direction for all)\ t`.
 

d.    `text(As)\ t->oo`,  `e^(-2t)=1/e^(2t)->0`

`=>8/e^(2t)->0\  text(and)`

`=>v=8-8/e^(2t)->8\ text(m/s)`
 

`:.\ text(As)\ \ t->oo,\ text(velocity approaches 8 m/s.)`

 

IMPORTANT: Use previous parts to inform this diagram. i.e. clearly show velocity was zero at  `t=0`  and the asymptote at  `v=8`. 
e.    

Calculus in the Physical World, 2UA 2011 HSC 7b Answer