SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, EXT1 C1 EQ-Bank 2

A particle is moving along the \(x\)-axis where its displacement, in metres from the origin, after \(t\) seconds, is given by:

\(x=t^3-5t^2+8t-6\).

  1. Determine the times when the particle is at rest.   (2 marks)

  2. Find and describe the velocity of the particle at the time where there is no accelerating force acting on it.  (2 marks)

Show Answers Only

i.    \(t=\dfrac{4}{3}\ \text{s  or}\  \ t=2\ \text{s}\)

ii.  \(\dot{x}=-\dfrac{1}{3}\ \text{m/s} \ \text{(towards the left)}\)

Show Worked Solution

i.    \(x=t^3-5t^2+8t-6\)

\(\dot{x}=3 t^2-10 t+8\)

\(\ddot{x}=6 t-10\)
 

\(\text {Particle at rest}\ \Rightarrow \ \dot{x}=0:\)

\(3 t^2-10 t+8\) \(=0\)  
\((3t-4)(t-2)\) \(=0\)  

 
\(t=\dfrac{4}{3}\ \text{s  or}\  \ t=2\ \text{s}\)
 

ii.  \(\text {No acceleration}\ \Rightarrow\ \ddot{x}=0\)

\(6 t-10=0\ \Rightarrow\ \ t=\dfrac{5}{3}\ \text{s}\)

\(\text{Find}\ \dot{x}\ \text{when}\ \ t=\dfrac{5}{3}: \)

  \(\dot{x}\) \( =3\left(\dfrac{5}{3}\right)^2-10\left(\dfrac{5}{3}\right)+8\)
    \(=\dfrac{25}{3}-\dfrac{50}{3}+\dfrac{24}{3} \)
    \( =-\dfrac{1}{3}\ \text{m/s}\)

 
\(\therefore\ \text{No accelerating force when}\ \ \dot{x}=-\dfrac{1}{3}\ \text{m/s} \ \text{(towards the left)}\)

Calculus, EXT1 C1 EQ-Bank 1

A particle is moving along the `x`-axis with a velocity, `dotx`, in metres per second at  `t`  seconds, is given by the function

`dotx = sqrt(5t+4t^2-t^3)`

Find the acceleration of the particle when  `t=3`.  (2 marks)

Show Answers Only

`sqrt6/12\ \ text(ms)^(-2)`

Show Worked Solution
`dotx` `=(5t+4t^2-t^3)^(1/2)`  
`ddot x` `=d/(dt)(dotx)`  
  `=1/2(5t+4t^2-t^3)^(-1/2) xx (5+8t-3t^2)`  

 
`text(When)\ \ t=3,`

`ddotx` `=1/2 (15+36-27)^(-1/2) xx (5 + 24-27)`  
  `=2/(2sqrt24)`  
  `=sqrt6/12\ \ text(ms)^(-2)`  

Calculus, EXT1* C1 2019 HSC 10 MC

A particle is moving along a straight line with displacement `x` at time `t`.

The particle is stationary when  `t = 11`  and when  `t = 13`.

Which of the following MUST be true in this case?

A.  The particle changes direction at some time between  `t = 11`  and  `t = 13`.
B.   The displacement function of the particle has a stationary point at some time between  `t = 11`  and  `t = 13`.
C.   The acceleration of the particle is 0 at some time between  `t = 11`  and  `t = 13`.
D.   The acceleration function of the particle has a stationary point at some time between  `t = 11`  and  `t = 13`.
Show Answers Only

`C`

Show Worked Solution

`text(Given)\ \ v = 0\ \ text(at)\ \ t = 11 and t = 13,`

♦ Mean mark 37%.

`text(Consider option)\ C:`

`text(If)\ v\ text(remains)\ 0, \ a = 0.`

`text(If)\ v\ text(increases)\ (a > 0 )\ text(or decreases)\ (a < 0)\ text(after)\ \ t = 11,`

`text(it must then decrease)\ (a < 0)\ text(or increase)\ (a > 0)`

`text(respectively so that)\ \ v = 0\ \ text(when)\ \ t = 13.`

`text(In each scenario,)\ a = 0\ text(between)\ \ t = 11 and t = 13.`

`=>  C`

Calculus, EXT1* C1 2017 HSC 15c

Two particles move along the `x`-axis.

When  `t = 0`, particle `P_1` is at the origin and moving with velocity 3.

For  `t >= 0`, particle `P_1` has acceleration given by  `a_1 = 6t + e^(-t)`.

  1. Show that the velocity of particle `P_1` is given by  `v_1 = 3t^2 + 4-e^(-t)`  (2 marks)

When  `t = 0`, particle `P_2` is also at the origin.

For  `t >= 0`, particle `P_2` has velocity given by  `v_2 = 6t + 1-e^(-t)`.

  1. When do the two particles have the same velocity?  (2 marks)

  2. Show that the two particles do not meet for  `t > 0`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t = 1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `a_1` `= 6t + e^(-t)`
  `v_1` `= int a_1\ dt`
    `= int 6t + e^(-t)\ dt`
    `= 3t^2-e^(-t) + c`

 

`text(When)\ t = 0,\ v_1 = 3`

`3` `= 0-1 + c`
`c` `= 4`
`:. v_1` `= 3t^2 + 4-e^(-t) …\ text(as required)`

 

ii.  `v_2 = 6t + 1-e^(-t)`

`text(Find)\ \ t\ \ text(when)\ \ v_1 = v_2`

`3t^2 + 4-e^(-t)` `= 6t + 1-e^(-t)`
`3t^2-6t + 3` `= 0`
`t^2-2t + 1` `= 0`
`(t-1)^2` `= 0`
`:. t` `=1`

 

iii.   `x_1` `= int v_1\ dt`
    `= int 3t^2 + 4-e^(-t)\ dt`
    `= t^3 + 4t + e^(-t) + c`

 

`text(When)\ \ t = 0,\ \ x_1 = 0`

♦ Mean mark (iii) 39%.
`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_1` `= t^3 + 4t + e^(-t)-1`

 

`x_2` `= int 6t + 1-e^(-t)\ dt`
  `= 3t^2 + t + e^(-t) + c`

 
`text(When)\ \ t = 0,\ \ x_2 = 0`

`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_2` `= 3t^2 + t + e^(-t)-1`

 

`text(Find)\ \ t\ \ text(when)\ \ x_1 = x_2`

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

 

`text(S)text(ince)\ \ Delta < 0\ \ text(for)\ \ t^2-3t + 3`

`=>\ text(No real solution)`

 

`:.\ text(The particles do not meet)`

`(x_1 != x_2)\ \ text(for)\ \ t > 0.`

Calculus, EXT1* C1 2004 HSC 9b

A particle moves along the `x`-axis. Initially it is at rest at the origin. The graph shows the acceleration, `a`, of the particle as a function of time  `t`  for  `0 ≤ t ≤ 5`.
 

2004 9b
 

  1. Write down the time at which the velocity of the particle is a maximum.  (1 marks)

  2. At what time during the interval  `0 ≤ t ≤ 5`  is the particle furthest from the origin? Give brief reasons for your answer.  (2 marks)

Show Answers Only
  1. `2`
  2. `4`
Show Worked Solution

i.   `text(The velocity of the particle increases until)`

`t = 2\ \ text{(when}\ \ a=0text{)}.` 

`:.\ text(Maximum velocity occurs at)\ \ t = 2.`

 

ii.  `text(It is furthest from the origin when its positive velocity)`

`text(slows and becomes zero. This occurs when the “net” area)`

`text(under the curve is zero).`

`text(The area under the curve and above the)\ x text(-axis)`

`text(between)\ \ t=0 and 2,\ text(is equal to the area above)`

`text(the curve and below the)\ x text(-axis between)\ \ t=2 and 4.`

`:.\ text(Furthest from origin when)\ \ t=4.`

Calculus, EXT1* C1 2007 HSC 5b

A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by

`v = (2t)/(16 + t^2).`

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

  2. Find an expression for the acceleration of the particle.  (2 marks)

  3. Find the time when the acceleration of the particle is zero.  (1 mark)

  4. Find the position of the particle when `t = 4`.  (3 marks)

Show Answers Only
  1. `0`
  2. `{2(16 – t^2)}/(16 + t^2)^2`
  3. `4\ text(seconds)`
  4. `log_e 2\ \ text(metres)`
Show Worked Solution

i.  `v = (2t)/(16 + t^2)`

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

`:.\ text(Initial velocity is 0.)`

 

ii.  `a = d/(dt) ((2t)/(16 + t^2))`
 

`text(Using quotient rule)`

`u` `= 2t` `v` `= 16 + t^2`
`u prime` `= 2` `v prime` `= 2t`
`(dv)/(dt)` `= (u prime v – uv prime)/v^2`
  `= {2(16 + t^2) – 2t * 2t}/(16 + t^2)^2`
  `= (32 + 2t^2 – 4t^2)/(16 + t^2)^2`
  `= {2(16 – t^2)}/(16 + t^2)^2`

 

iii.  `text(Find)\ t\ text(when)\ (dv)/(dt) = 0`

`{2 (16 – t^2)}/(16 + t^2)^2` `= 0`
`2 (16 – t^2)` `= 0`
`t^2` `= 16`
`t` `= 4\ ,\ t >= 0`

 

`:.\ text(The acceleration is zero when)`

`t = 4\ text(seconds.)`

 

iv.  `v = (2t)/(16 + t^2)`

`x` `= int v\ dt`
  `= int (2t)/(16 + t^2)`
  `= log_e (16 + t^2) + c`

 

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

`0 = log_e (16 + 0) + c`

`c = -log_e 16`

`:. x = log_e(16 + t^2) – log_e 16`
 

`text(When)\ t = 4,`

`x` `= log_e (16 + 4^2) – log_e 16`
  `= log_e 32 – log_e 16`
  `= log_e (32/16)`
  `= log_e 2\ \ text(metres)`

 

`:.\ text(When)\ t = 4\ , \ text(the position of the)`

`text(particle is)\ log_e 2 \ text(metres.)`

Calculus, EXT1* C1 2015 HSC 14a

In a theme park ride, a chair is released from a height of  `110`  metres and falls vertically. Magnetic brakes are applied when the velocity of the chair reaches  `text(−37)`  metres per second.
 

2015 2ua 14a
 

The height of the chair at time `t` seconds is `x` metres. The acceleration of the chair is given by   `ddot x = −10`. At the release point,  `t = 0, x = 110 and dot x = 0`.

  1. Using calculus, show that  `x = -5t^2 + 110`.  (2 marks)

  2. How far has the chair fallen when the magnetic brakes are applied?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `68.45\ text(m)`
Show Worked Solution

i.  `text(Show)\ \ x = -5t^2 + 110`

`ddot x` `= -10`
`dot x` `= int ddot x\ dt`
  `= int -10\ dt`
  `= -10t + c`

 
`text(When)\ \ t = 0,\ dot x = 0`

`:.\ 0` `= -10 (0) + c`
`c` `= 0`
`dot x` `= -10t`
`x` `= int dot x\ dt`
  `= int -10t\ dt`
  `= -5t^2 + c`

 
`text(When)\ \ t = 0,\ x = 110`

`:.\ 110` `= -5 (0^2) + c`
`c` `= 110`

 
`:.\ x = -5t^2 + 110\ \ text(…  as required.)`
 

ii.  `text(Find)\ \t\ \text(when)\ \ dot x = -37`

`-37` `= -10t`
`t` `= 3.7\ \ text(seconds)`

 
`text(When)\ \ t = 3.7`

`x` `= -5 (3.7^2) + 110`
  `= -68.45 + 110`
  `= 41.55`

 
`:.\ text(Distance the chair has fallen)`

`= 110 – 41.55`

`= 68.45\ text(m)`

Calculus, EXT1* C1 2005 HSC 9a

A particle is initially at rest at the origin. Its acceleration as a function of time, `t`, is given by

`ddot x = 4sin2t`

  1. Show that the velocity of the particle is given by  `dot x = 2 − 2\ cos\ 2t`.  (2 marks)

  2. Sketch the graph of the velocity for  `0 ≤ t ≤ 2π`  AND determine the time at which the particle first comes to rest after  `t = 0`.  (3 marks)

  3. Find the distance travelled by the particle between  `t = 0`  and the time at which the particle first comes to rest after  `t = 0`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2pi\ \ text(units)`
Show Worked Solution

i.   `text(Show)\ \ dot x = 2 − 2\ cos\ 2t`

`ddot x` `= 4\ sin\ 2t`
`dot x` `= int ddot x\ dt`
  `= int4\ sin\ 2t\ dt`
  `= −2\ cos\ 2t + c`

 
`text(When)\ t = 0, \ x = 0`

`0 = −2\ cos\ 0 + c`

`c = 2`

`:.dot x = 2 − 2\ cos\ 2t\ \ …text(as required)`

 

ii.  `text(Considering the range)`

`−1` `≤ \ \ \ cos\ 2t` `≤ 1`
`−2` `≤ \ \ \ 2\ cos\ 2t`  `≤ 2` 
`0` `≤ 2 − 2\ cos\ 2t` `≤ 4` 

 
`text(Period) = (2pi)/n = (2pi)/2 = pi`

Calculus in the Physical World, 2UA 2005 HSC 9a Answer

`text(After)\ t = 0,\ text(the particle next comes)`

`text(to rest at)\ t = pi.`

 

iii.  `text(Distance travelled)`

`= int_0^pi dot x\ dt`

`= int_0^pi 2 − 2\ cos\ 2t\ dt`

`= [2t − sin\ 2t]_0^pi`

`= [(2pi − sin\ 2pi) − (0 − sin\ 0)]`

`= 2pi\ \ text(units)`

Calculus, EXT1* C1 2005 HSC 7b

Calculus in the Physical World, 2UA 2005 HSC 7b
 

The graph shows the velocity, `(dx)/(dt)`, of a particle as a function of time. Initially the particle is at the origin. 

  1. At what time is the displacement, `x`, from the origin a maximum?  (1 mark)

  2. At what time does the particle return to the origin? Justify your answer.  (2 marks)

  3. Draw a sketch of the acceleration,  `(d^2x)/(dt^2)`, as a function of time for  `0 ≤ t ≤ 6`.  (2 marks)

Show Answers Only
  1. `2`
  2. `4`
  3. `text(See Worked Solutions.)`
Show Worked Solution

i.   `text(Maximum displacement in graph when)`

`(dx)/(dt)` `= 0`
`:.t` `= 2`

 

ii.  `text(Particle returns to the origin when)\ t = 4.`

`text(The displacement can be calculated by the)`

`text(net area below the curve and since the)`

`text(area above the curve between)\ t = 0\ text(and)\ t = 2`

`text(is equal to the area below the curve between)`

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

`text{the initial displacement (i.e. the origin).}`

 

iii.   Calculus in the Physical World, 2UA 2005 HSC 7b Answer

Calculus, EXT1* C1 2013 HSC 10 MC

A particle is moving along the  `x`-axis. The displacement of the particle at time  `t`  seconds is  `x`  metres.

At a certain time,  `dot x = -3\ text(ms)^(-1)`  and  `ddot x = 2\ text(ms)^(-2)`.

Which statement describes the motion of the particle at that time?

  1. The particle is moving to the right with increasing speed.
  2. The particle is moving to the left with increasing speed.
  3. The particle is moving to the right with decreasing speed.
  4. The particle is moving to the left with decreasing speed.
Show Answers Only

`D`

Show Worked Solution
♦♦ Mean mark 26%.

`text(S)text(ince)\ dot x = -3 text(ms)`-1

`=>\ text(Particle is moving to the left)`

`text(S)text(ince)\ ddot x = 2 text(ms)`-2

`=>\ text(Acceleration is to the right, against)`

`text(the particle’s motion)`

`:.\ text(Speed is decreasing)`

`=>  D`

Calculus, EXT1* C1 2010 HSC 7a

The acceleration of a particle is given by

`ddotx=4cos2t`,

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

Initially the particle is at the origin with a velocity of  `text(1 ms)^(–1)`.

  1. Show that the velocity of the particle is given by

     

      `dotx=2sin2t+1`.    (2 marks)

  2. Find the time when the particle first comes to rest.    (2 marks)

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

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(7pi)/12\ text(seconds)`
  3. `x=1-cos2t+t`
Show Worked Solution
i.   `text(Show)\ dotx` `=2sin2t+1`
`dotx` `=intddotx\ dt`
  `=int4cos2t\ dt`
  `=2sin2t+c`

 
`text(When)\ t=0, \ \ dotx=1\ \ text{(given)}`

`1=2sin0+c`

`c=1`

 
`:. dotx=2sin2t+1 \ \ \ text(… as required)`
 

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

`2sin2t+1` `=0`
`sin2t` `=-1/2`

 
`=>sin theta=1/2\ text(when)\ theta=pi/6`

`text(S)text(ince)\ \ sin theta\ \ text(is negative in 3rd and 4th quadrants)`

`2t` `=pi + pi/6`
`2t` `=(7pi)/6`
`t` `=(7pi)/12`

 
`:.\ text(Particle first comes to rest at)\ t=(7pi)/12\ text(seconds)`
 

iii.    `x` `=intdotx\ dt`
  `=int(2sin2t+1)\ dt`
  `=t-cos2t+c`

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

`0=0-cos0+c`

`c=1`

 
`:. x=t-cos2t+1`

Calculus, EXT1* C1 2013 HSC 14a

The velocity of a particle moving along the `x`-axis is given by  `dotx=10-2t`, where `x` is the displacement from the origin in metres and `t` is the time in seconds. Initially the particle is 5 metres to the right of the origin.

  1. Show that the acceleration of the particle is constant.  (1 mark)

  2. Find the time when the particle is at rest.  (1 mark)

  3. Show that the position of the particle after 7 seconds is 26 metres to the right of the origin.  (2 marks)

  4. Find the distance travelled by the particle during the first 7 seconds.   (2 marks)

Show Answers Only
  1.  `ddotx=-2\ \ text{(constant)}`
  2. `t=5\ text(seconds)`
  3. `text{Proof (See Worked Solutions)}`
  4. `29\ text(metres)`
Show Worked Solutions

i.   `dotx=10-2t`

`text(Acceleration)=ddotx=d/(dx)dotx=-2`

 
`:.\ text(The acceleration is constant.)`

 

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

`10-2t=0\ text(when)\ t=5`

 
`:.\ text(Particle comes to rest after 5 seconds.)`

 

 iii. `text(Show)\  x=26\ text(when)\ t=7`

`x` `=int dotx\ dt`
  `=int (10-2t)\ dt`
  `=10t -t^2+c`

 

`text(Given)\  t=0\ text(when)\ x=5`

`=> 5` `=10(0)-0^2+c`
`c` `=5`

 
`:. x=10t-t^2+5`
 

`text(At)\ \ t=7`

`x` `=10(7)-7^2+5`
  `=70-49+5`
  `=26`

 
`:.\ text(The particle is 26 metres to the right of the origin)`

`text(after 7 seconds … as required)`

 

 iv.   `text(Find the distance travelled in the first 7 seconds:)`

♦ Mean mark 37%.
IMPORTANT: Students can also draw a number line that shows where a particle starts from, changes direction and finishes in order to reduce errors.

`text(At)\  t=5`,

`x` `=10(5)-5^2+5`
  `=50-25+5`
  `=30\ text(m)`

 
`=>\ text{The particle travels 25 m to the right}\ (x=5\ text{to}\ 30)`

`text{then 4 m to the left}\  (x=30\ text{to}\ 26)`
 

`:.\ text(The total distance travelled in 7 seconds)`

`=25+4`

`=29\ text(m)`