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Calculus, 2ADV C1 EQ-Bank 8

The displacement \(x\) metres from the origin at time, \(t\) seconds, of a particle travelling in a straight line is given by

\(x=t^3-9 t^2+9 t, \quad t \geqslant 0\)

  1. Find the time(s) when the particle is at the origin.   (2 marks)
  2. On the graph below, sketch the displacement, \(x\) metres, with respect to time \(t\).   (2 marks)
     
       

  3. Find the velocity of the particle when  \(t=2\).   (2 marks)

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a.  \(\text{Particle at origin when}\ \ t=0, t=3.\)

b.
       
 

c.   \(\dot{x}=-15\  \text{m s}^{-1}\)

Show Worked Solution

a.    \(x\) \(=t^3-9 t^2+9 t\)
    \(=t\left(t^2-9 t+9\right)\)
    \(=t(t-3)^2\)

 
\(\text{Particle at origin when}\ \ t=0, t=3.\)

 
b.
       
 

c.    \(x=t^3-9 t^2+9 t\)

\(\dot{x}= \dfrac{dx}{dt} = 3 t^2-18 t+9\)

\(\text {When } t=2:\)

\(\dot{x}=3 \times 2^2-18 \times 2+9=-15\  \text{m s}^{-1}\)

Calculus, 2ADV C1 EQ-Bank 5

A magpie plague hit Raymond Terrace this year but was eventually brought under control. A bird researcher estimated that the magpie population \(M\), in hundreds, \(t\) months after 1st January, was given by  \(M=7+20t-3t^2\)

  1. Find the magpie population on 1st March.   (1 mark)
  2. At what rate was the population changing at this time?   (1 mark)
  3. In what month does the magpie population start to decrease?   (2 marks)
Show Answers Only

a.    \(\text{Population}\ =35 \times 100=3500\)

b.   \(M\ \text{is increasing at 800 per month}\)

c.   \(\text{Population starts to decrease in May.}\)

Show Worked Solution

a.    \(\text{Find}\ M\ \text{when}\ \ t=2:\)

\(M=7+20 \times 2-3 \times 2^2 = 35\)

\(\therefore \text{Population}\ =35 \times 100=3500\)
 

b.    \(M=7+20t-3t^2\)

\(\dfrac{dM}{dt}=20-6t\)

\(\text{Find}\ \dfrac{dM}{dt}\ \text{when}\ \ t=2: \)

\(\dfrac{dM}{dt}=20-6 \times 2 = 8\)

\(\therefore M\ \text{is increasing at 800 per month}\)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dM}{dt}=0: \)

\(\dfrac{dM}{dt}=20-6t = 0\ \ \Rightarrow \ t= 3\ \dfrac{1}{3}\)

\(\dfrac{dM}{dt}<0\ \ \text{when}\ \ t>3\ \dfrac{1}{3} \)

\(\therefore\ \text{Population starts to decrease in May.}\)

Calculus, 2ADV C1 EQ-Bank 2 MC

The displacement of a particle is given by  \(x=3t^{3}-6t^{2}-15\) . The acceleration is zero at:

  1. \(t=\dfrac{2}{3}\)
  2. \(t=\dfrac{4}{3}\)
  3. \(t=\dfrac{5}{2}\)
  4. \(\text{never}\)
Show Answers Only

\(A\)

Show Worked Solution

\(x=3t^{3}-6t^{2}-15\)

\(v=9t^{2}-12t\)

\(a=18t-12\)

\(\text{Find}\ t\ \text{when}\ \ a=0:\)

\(18t-12=0\ \ \Rightarrow\ \ t=\dfrac{2}{3} \)

\(\Rightarrow A\)

Calculus, 2ADV C1 SM-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3 - t^2 - 3t + 11`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 2`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only
  1.  `17\ text(ms)^(−1)`
  2.  `(1 + sqrt19)/6`
Show Worked Solution

i.   `x =2t^3 – t^2 – 3t + 11` 

`v = (dx)/(dt) = 6t^2 – 2t – 3`

 
`text(When)\ t = 2,`

`v` `= 6 xx 2^2 – 2 · 2 – 3`
  `= 17\ text(ms)^(−1)`

 

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

`6t^2 – 2t – 3` `= 0`
`:. t` `= (2 ±sqrt((−2)^2 – 4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`

Calculus, 2ADV C1 SM-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = sqrt(20t - 2t^2)`  metres per second

Find the acceleration of the particle when  `t = 4`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

`sqrt3/6\ \ text(ms)^(−2)`

Show Worked Solution

`v = sqrt(20t – 2t^2)`

`alpha` `= (dv)/(dt)`
  `= 1/2 · (20t – 2t^2)^(−1/2) · (20 – 4t)`

 
`text(When)\ \ t = 4,`

`alpha` `= 1/2(20 · 4 – 2 · 4^2)^(−1/2)(20 – 16)`
  `= 2/(sqrt48)`
  `= 2/(4sqrt3) xx sqrt3/sqrt3`
  `= sqrt3/6\ \ text(ms)^(−2)`

Calculus, 2ADV C1 2008 HSC 6b

The graph shows the velocity of a particle,  `v`  metres per second, as a function of time,  `t`  seconds.

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

  2. When is the velocity of the particle equal to zero?    (1 mark)

  3. When is the acceleration of the particle equal to zero?    (1 mark)

Show Answers Only
  1. `20\ text(m/s)`
  2. `t=10\ text(seconds)`
  3. `t=6\ text(seconds)`
Show Worked Solution

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

`v=20\ \ text(m/s)`
 

ii.    `text(Particle comes to rest at)\  t=10\ text{seconds  (from graph)}`

 

iii.  `text(Acceleration is zero when)\ t=6\ text{seconds  (from graph)}`

Calculus, 2ADV C1 2018 HSC 12d

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

`x = t^3/3 - 2t^2 + 3t,`

where `x` is the displacement from the origin in metres and `t` is the time in seconds, for `t >= 0`.

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

  2. At which times is the particle stationary?  (2 marks)

  3. Find the position of the particle when the acceleration is zero.  (2 marks)

Show Answers Only
  1. `3\ text(ms)^(-1)`
  2. `t = 1 or 3\ text(seconds)`
  3. `2/3\ text(m)`
Show Worked Solution

i.    `x = t^3/3 – 2t^2 + 3t`

`v = (dx)/(dt) = t^2 – 4t + 3`
 

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

`v` `= 0 – 0 + 3`
  `= 3\ text(ms)^(-1)`

 

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

`t^2 – 4t + 3 = 0`

`(t – 3) (t – 1) = 0`

`t = 1 or 3\ text(seconds)`
 

iii.  `a = (dv)/(dt) = 2t – 4`
 

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

`2t – 4` `= 0`
`t` `= 2`
`x(2)` `= 2^3/3 – 2(2^2) + 3(2)`
  `= 8/3 – 8 + 6`
  `= 2/3`

Calculus, 2ADV C1 2006 HSC 8a

A particle is moving in a straight line. Its displacement, `x` metres, from the origin, `O`, at time `t` seconds, where  `t ≥ 0`, is given by  `x = 1 - 7/(t + 4)`.

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

  2. Find the velocity of the particle as it passes through the origin.  (3 marks)

  3. Show that the acceleration of the particle is always negative.  (1 mark)

  4. Sketch the graph of the displacement of the particle as a function of time.  (2 marks)

Show Answers Only
  1. `text(–3/4 m)`
  2. `1/7\ text(ms)^-1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4.  
Show Worked Solution

i.   `x = 1 – 7/(t + 4)`
 

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

`x` `= 1 – 7/4`
  `= -3/4 \ \ text(m)`

 
`:.\ text(Initial displacement is)\ 3/4\ text(metres to)`

`text(the left of the origin.)`

 

ii.  `x = 1 – 7/(t+4) = 1 – 7(t + 4)^-1`

`dot x` `= (-1)  -7(t + 4)^-2 xx d/(dt)(t + 4)`
  `= 7 (t + 4)^-2 xx 1`
  `= 7/(t + 4)^2`

 

`text(Find)\ t\ text(when)\ x = 0`

`0` `= 1 – 7/(t + 4)`
`7/(t + 4)` `= 1`
`7` `= (t + 4)`
`t` `= 3`

 

`text(When)\ t = 3`

`dot x` `= 7/(3 + 4)^2`
  `= 1/7\ text(ms)^-1`

 

`:.\ text(The velocity of the particle as it passes)`

`text(through the origin is)\ 1/7\ text(ms)^-1.`

 

iii.  `dot x` `= 7(t + 4)^-2`
`ddot x` `= (d dot x)/(dt) = -14 (t +4)^-3`

 
`text(Given)\ t >= 0`

`=>  (t + 4)^-3 >= 0`

`=> -14 (t + 4)^-3 <= 0`

`:. ddot x\ text(is always negative.)`
 

(iv)  2UA HSC 2006 8a

Calculus, 2ADV C1 2014 HSC 13c

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

 `x = t - 1/(1 + t)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1`
Show Worked Solution
i.    `x` `= t\ – 1/(1 + t)`
    `= t\ – (1 + t)^(-1)`

 

`dot x` `= 1\ – (-1) (1 + t)^(-2)`
  `= 1 + 1/((1 + t)^2)`

 

`ddot x` `= -2(1 + t)^(-3)`
  `= – 2/((1 + t)^3)`

 
`text(S)text(ince)\ \ t >= 0,`

`=> -2/((1 + t)^3) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

ii.    `text(Velocity)\ (dot x) = 1 + 1/((1 + t)^2)`

 
`text(As)\ t -> oo,\ 1/((1 + t)^2) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 1`

Calculus, 2ADV C1 2014 HSC 9 MC

The graph shows the displacement  `x`  of a particle moving along a straight line as a function of time  `t`.

2014 9 mc

 Which statement describes the motion of the particle at the point  `P`? 

  1. The velocity is negative and the acceleration is positive.
  2. The velocity is negative and the acceleration is negative.
  3. The velocity is positive and the acceleration is positive.
  4. The velocity is positive and the acceleration is negative.
Show Answers Only

`A`

Show Worked Solution

`text(At)\ P,\ text(the particle is moving back towards)\ O`

`text{after hitting a max (positive) displacement}`

`:.\ text(Velocity is negative.)`

`text(Its displacement hits a minimum just after)\ P`

`text(and increases again.)`

`:.\ text{Acceleration is working against (negative) velocity}`

`text(and must be positive.)`

`=>  A`