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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)

Show Answers Only

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 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`