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Vectors, SPEC2 2024 VCAA 16 MC

Particle 1 has position vector  \({\underset{\sim}{r}}_1(t)=\cos (t) \underset{\sim}{ i }+\sin (t) \underset{\sim}{ j }+\sqrt{\sin (2 t)} \underset{\sim}{ k }\)  and Particle 2 has position vector  \({\underset{\sim}{r}}_2(t)=\sin (t) \underset{\sim}{ i }+\cos (t) \underset{\sim}{ j }+\sqrt{\sin (2 t)} \underset{\sim}{ k }\), where \(t\) is measured in seconds and  \(t \in\left(0, \dfrac{\pi}{2}\right)\).

The number of times the velocity of Particle 1 is perpendicular to the position vector  \({\underset{\sim}{r}}_2(t)\) during the first \(\dfrac{\pi}{2}\) seconds is

  1. \(1\)
  2. \(2\)
  3. \(3\)
  4. \(4\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{By CAS:}\)

\({\underset{\sim}{v}}_1(t)=\dfrac{d}{dt} \left({\underset{\sim}{r}}_1(t)\right)=-\sin (t)\underset{\sim}{i}+\cos (t) \underset{\sim}{j}+\dfrac{\cos (2 t)}{\sqrt{\sin (2 t)}}\, \underset{\sim}{k}\)

\(\text{Find when} \ {\underset{\sim}{v}}_1(t) \perp {\underset{\sim}{r}}_2(t)\)

\(\text{Solve}\ \ {\underset{\sim}{v}}_{1} (t) \cdot {\underset{\sim}{r}}_{2} (t)=0\ \ \text{for}\ \ t \in\left(0, \dfrac{\pi}{2}\right):\)

\({\underset{\sim}{v}}_1(t) \cdot {\underset{\sim}{r}}_2(t)=-\sin ^2(t)+\cos ^2(t)+\cos (2 t)=2 \cos (2 t)\)

\(2 t=\dfrac{\pi}{2} \ \Rightarrow \ t=\dfrac{\pi}{4}\)

\(\Rightarrow A\)

♦ Mean mark 43%.

Vectors, SPEC2 2022 VCAA 13 MC

The acceleration of a body moving in a plane is given by  \(\underset{\sim}{\ddot{\text{r}}}(t)=\sin(t)\underset{\sim}{\text{i}}+2 \cos(t)\underset{\sim}{\text{j}}\), where  \(t \ge 0\).

Given that  \(\underset{\sim}{\dot{\text{r}}}(0)=2\underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\), the velocity of the body at time \(t, \underset{\sim}{\dot{\text{r}}}(t)\), is given by

  1. \(-\cos (t) \underset{\sim}{\text{i}}+2 \sin (t) \underset{\sim}{\text{j}}\)
  2. \((3-\cos (t)) \underset{\sim}{\text{i}}+(2 \sin (t)+1) \underset{\sim}{\text{j}}\)
  3. \((1+\cos (t)) \underset{\sim}{\text{i}}+(2\sin (t)+1) \underset{\sim}{\text{j}}\)
  4. \((2+\sin (t)) \underset{\sim}{\text{i}}+(2\cos (t)-1) \underset{\sim}{\text{j}}\)
  5. \((1+\cos (t)) \underset{\sim}{\text{i}}+(1-2\sin (t)) \underset{\sim}{\text{j}}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\underset{\sim}{\ddot{r}}(t)=\sin (t) \underset{\sim}{i}+2 \cos (t) \underset{\sim}{j}\)

\(\underset{\sim}{\dot{r}}(t)=\displaystyle{\int} \ddot{r}(t) d t=-\cos (t) \underset{\sim}{i}+2 \sin (t) \underset{\sim}{j}+\underset{\sim}{c}\)

\(\text {When } t=0,\  \dot{r}(t)=2 \underset{\sim}{i}+\underset{\sim}{j}:\)

\begin{aligned}
2 \underset{\sim}{i}+\underset{\sim}{j} & =-\cos (0) \underset{\sim}{i}+0+\underset{\sim}{c} \\
\underset{\sim}{c} & =3\underset{\sim}{i}+\underset{\sim}{j}
\end{aligned}

\begin{aligned}
\therefore \dot{r}(t) & =-\cos(t)\underset{\sim}{i}+2 \sin (t) \underset{\sim}{j}+3 \underset{\sim}{i}+\underset{\sim}{j}\\
& =(3-\cos (t)) \underset{\sim}{i}+(2 \sin (t)+1) \underset{\sim}{j}
\end{aligned}

\(=>B\)

Vectors, SPEC1 2023 VCAA 10

The position vector of a particle at time \(t\) seconds is given by

\(\underset{\sim}{\text{r}}(t)=\big{(}5-6 \ \sin ^2(t) \big{)} \underset{\sim}{\text{i}}+(1+6 \ \sin (t) \cos (t)) \underset{\sim}{\text{j}}\), where \(t \geq 0\).

  1. Write \(5-6\, \sin ^2(t)\) in the form \(\alpha+\beta\, \cos (2 t)\), where \(\alpha, \beta \in Z^{+}\).  (1 mark)
  1. Show that the Cartesian equation of the path of the particle is \((x-2)^2+(y-1)^2=9.\)  (2 marks)
  1. The particle is at point \(A\) when \(t=0\) and at point \(B\) when \(t=a\), where \(a\) is a positive real constant.
  2. If the distance travelled along the curve from \(A\) to \(B\) is \(\dfrac{3 \pi}{4}\), find \(a\).   (1 mark)
  1. Find all values of \(t\) for which the position vector of the particle, \(\underset{\sim}{\text{r}}(t)\), is perpendicular to its velocity vector, \(\underset{\sim}{\dot{\text{r}}}(t)\).   (2 marks)
Show Answers Only

a.    \(2+3\, \cos (2 t) \)

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

c.    \(a=\dfrac{\pi}{8}\)

d.    \(t =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) \)

Show Worked Solution

a.  \(5-6\, \sin ^2(t)=5-6 \times \dfrac{1}{2}(1-\cos (2 t))\)

\(\ \ \ \quad \quad \quad \quad \quad \quad \begin{aligned}
& =5-3+3\, \cos (2 t) \\
& =2+3\, \cos (2 t)
\end{aligned}\)
 

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

 
c.  \(\text { Motion is circular, centre }(2,1) \text {, radius }=3\)

\begin{aligned}
\text { Arc length } & = r \theta \\
3 \theta & =\dfrac{3 \pi}{4} \\
\theta & =\dfrac{\pi}{4}
\end{aligned}

\(\therefore a=\dfrac{\pi}{8}\)
 

d.    \(\underset{\sim}{r}=(2+3\, \cos (2 t)) \underset{\sim}{i}+(1+3\, \sin (2 t)) \underset{\sim}{j}\)

\(\underset{\sim}{\dot{r}}=-6\, \sin (2 t) \underset{\sim}{i}+6\, \cos (2 t) \underset{\sim}{j}\)

\(\text { Find } t \text { when } \underset{\sim}{r} \cdot \underset{\sim}{\dot{r}}=0 \text { : }\)

\begin{aligned}
\underset{\sim}{r} \cdot \underset{\sim}{\dot{r}} & =6\left(\begin{array}{l}
2+3\, \cos (2 t) \\
1+3\, \sin (2 t)
\end{array}\right)\left(\begin{array}{l}
-\sin (2 t) \\
\cos (2 t)
\end{array}\right) \\
0 &=-2\,\sin (2 t)-3\, \cos (2 t) \sin (2 t)+\cos (2 t)+3\, \cos (2 t) \sin (2 t)\\
0 &=-2\, \sin (2 t)+\cos (2 t)\\
2\, \sin (2 t) &=\cos (2 t)\\
\tan (2 t) & =\dfrac{1}{2} \\
2 t & =\tan ^{-1}\left(\dfrac{1}{2}\right)+k \pi \\
t & =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) 
\end{aligned}

Vectors, SPEC2 2020 VCAA 4

A pilot is performing at an air show. The position of her aeroplane at time `t` relative to a fixed origin `O` is given by 

   `underset~r_text(A) (t) = (450-150sin((pit)/6))underset~i + (400-200cos((pit)/6))underset~j`,

where  `underset~i`  is a unit vector in a horizontal direction, `underset~j`  is a unit vector vertically up, displacement components are measured in metres and time `t` is measured in seconds where  `t >= 0`.

  1. Find the maximum speed of the aeroplane. Give your answer in `text(ms)^(−1)`.  (3 marks)

  2.  i. Use  `underset~r_text(A)(t)`  to show that the cartesian equation of the path of the aeroplane is given by
  3.     `((x-450)^2)/(22\ 500) + ((y-400)^2)/(40\ 000) = 1`.  (2 marks)

  4. ii. Sketch the path of the aeroplane on the axes provided below. Label the position of the aeroplane when  `t = 0`, using coordinates, and use an arrow to show the direction of motion of the aeroplane.  (3 marks)

  5.   

         
     

A friend of the pilot launches an experimental jet-powered drone to take photographs of the air show. The position of the drone at time `t` relative to the fixed origin is given by  `underset~r_text(D)(t) = (30t)underset~i + (−t^2 + 40t)underset~j`, where `t` is in seconds and  `0 <= t <= 40, underset~i`  is a unit vector in the same horizontal direction, `underset~j`  is a unit vector vertically up, and displacement components are measured in metres.

  1. Sketch the path of the drone on the axes provided in part b.ii. Using coordinates, label the points where the path of the drone crosses the path of the aeroplane, correct to the nearest metre.  (3 marks)

  2. Determine whether the drone will make contact with the aeroplane. Give reasons for your answer.  (3 marks)

Show Answers Only

  1. `(100pi)/3 text(ms)^(−1)`
  2.  i. `text(See Worked Solutions)`
  3. ii.
  4.  

     
  5. `text(See Worked Solutions)`

Show Worked Solution

a.    `underset~r_text(A)′(t)` `= −25picos((pit)/60)underset~i + 100/3pisin((pit)/6)underset~j`
    `= (25pi)/3(−3cos((pit)/6) + 4sin((pit)/6))`
`|underset~r_text(A)′(t)|` `= (25pi)/3 sqrt(9cos^2((pit)/6) + 16sin^2((pit)/6))`
  `= (25pi)/3 sqrt(9 + 7sin^2((pit)/6))`
`:. |underset~r_text(A)′(t)|_text(max)` `= (25pi)/3 sqrt(9 + 7)`
  `= (100pi)/3\ text(ms)^(−1)`

 

b. i.   `x = 450-150 sin((pit)/6) \ => \ sin((pit)/6) = (450-x)/150`

`sin^2((pit)/6) = ((x-450)^2)/(22\ 500)`
 

`y = 400-200cos((pit)/6) \ => \ cos((pit)/6) = (400-y)/200`

`cos^2((pit)/6) = ((y-400)^2)/(40\ 000)`
 

`text(Using)\ \ sin^2theta + cos^2theta = 1:`

`((x-450)^2)/(22\ 500) + ((y-400)^2)/(40\ 000) = 1`

 

b. ii.   `text(When)\ x = 450, \ y = 200, 600`

`text(When)\ y = 400, \ x = 300, 600`

`text(As)\ \ t -> 1, \ x↓, \ y↑`

`:.\ text(Motion is clockwise.)`
 

 

c.   `x = 30t \ => \ t = x/30`

`y` `= −t^2 + 40t`
  `= −(x^2)/900 + 4/3 x`

 

 

d.   `text(The graph shows 2 points where the paths cross.)`

`text(Consider)\ (316, 310):`

`text(Drone passes through when)`

`t = 316/30 ~~ 10.53\ text(seconds)`

`text(Plane passes through when)`

`450-150sin((pit)/6) = 316 \ => \ t ~~ 12.85\ text{seconds (no contact)}`
 

`text(Similarly, consider)\ (600, 400):`

`text(Drone passes through when)`

`t = 600/30 = 20\ text(seconds)`

`text(Plane passes through when)`

`400-200cos((pit)/6) = 400 \ => \ t = 10\ text(or)\ ~~ 14.7\ text{seconds (no contact)}`
  

`:.\ text(Drone will not make contact with the plane.)`

Vectors, SPEC2 2020 VCAA 1

A particle moves in the  `x\ – y`  plane such that its position in terms of `x` and `y` metres at `t` seconds is given by the parametric equations

`x = 2sin(2t)`

`y = 3cos(t)`

where `t >= 0`

  1. Find the distance, in metres, of the particle from the origin when  `t = pi/6`.   (2 marks)

  2.   i. Express  `(dy)/(dx)`  in terms of `t` and, hence, find the equation of the tangent to the path of the particle at  `t = pi`  seconds.   (3 marks)

  3.  ii. Find the velocity, `underset ~ v`, in `text(ms)^(−1)`, of the particle when  `t = pi`.   (2 marks)

  4. iii. Find the magnitude of the acceleration, in `text(ms)^(−2)`, when  `t = pi`.   (2 marks)

  5. Find the time, in seconds, when the particle first passes through the origin.   (1 mark)

  6. Express the distance, `d` metres, travelled by the particle from  `t = 0`  to  `t = pi/6`  as a definite integral and find this distance correct to three decimal places.   (2 marks)

Show Answers Only
  1. `sqrt39/2\ text(metres)`
  2.   i. `y = −3`
  3.  ii. `4underset~i`
  4. iii. `3\ text(ms)^(−2)`
  5. `pi/2\ \ (text(1st time))`
  6. `1.804\ \ (text(to 3 d.p.))`
Show Worked Solution

a.   `text(At)\ \ t = pi/6,`

`x = 2sin\ pi/3 = sqrt3`

`y = 3cos\ pi/6 = (3sqrt3)/2`

`:.\ text(Distance)` `= sqrt((sqrt3)^2 + ((3sqrt3)/2)^2)`
  `= sqrt39/2\ text(metres)`

 

b.i.   `(dx)/(dt) = 4cos(2t),\ \ (dy)/(dt) = −3sin(t)`

`(dy)/(dx)` `= (dy)/(dt) · (dt)/(dx)`
  `= (−3sin(t))/(4cos(2t))`

 
`text(When)\ t = pi :`

`(dy)/(dx)` `= (−3sin(pi))/(4cos(2pi))=0`

 
`text(Equation of tangent where)\ \ m = 0,\ text(through)\ (0, −3):`

`y = −3`
  

b.ii.   `underset~r(t)` `= 2sin(2t)underset~i + 3cos(t)underset~j`
  `underset~v(t)` `= 4cos(2t)underset~i-3sin(t)underset~j`
  `underset~v(pi)` `= 4cos(2pi)underset~i-3sin(pi)underset~i`
    `= 4underset~i`

 

b.iii.  `underset~a(t) = −8sin(2t)underset~i-3cos(t)underset~j`

♦ Mean mark (b)(iii) 48%.
`underset~a(pi)` `= −8sin(2pi)underset~i-3cos(pi)underset~j`
  `= 3underset~j`
`|underset~a(pi)|` `= sqrt(0^2 + 3^2)`
  `= 3\ text(ms)^(−2)`

 

c.   `text(Find)\ t\ text(when)\ \ x = 2sin(2t) = 0\ \ text(and)\ \ y = 3cos(t) = 0:`

`t = pi/2\ \ (text(1st time))`

 

d.    `text(Distance)` `= int_0^(pi/6) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(pi/6) sqrt((4cos(2t))^2 + (−3sin(t))^2)\ dt`
    `~~ 1.804\ \ (text(to 3 d.p.))`

Vectors, SPEC1 2012 VCAA 9

The position of a particle at time  `t`  is given by

`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`

  1. Find the velocity of the particle at time  `t.`   (1 mark)

  2. Find the speed of the particle at time  `t = 1`  in the form  `(a sqrt b)/c`, where `a, b` and `c` are positive integers.   (2 marks)

  3. Show that at time  `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).`   (2 marks)

  4. Find the angle in terms of `pi`, between the vector  `-sqrt 3 underset ~i + underset ~j`  and the vector  `underset ~r (t)`  at time  `t = 0.`   (2 marks)

Show Answers Only
  1. `((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`
  2. `(4 sqrt 6)/3`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(7 pi)/12`
Show Worked Solution

a.  `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`

  `underset ~v(t)` `=dot underset ~r(t)`
    `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j`
    `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`

 

b.    `underset ~v(1)` `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j`
    `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j`
  `|\ underset ~v(1)\ |` `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)`
    `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)`
    `= sqrt(8/3 + 8)`
    `= sqrt(32/3)`
    `= (4sqrt2)/sqrt3`
    `= (4 sqrt 6)/3`

♦ Mean mark part (c) 41%.

c.    `(dy)/(dt)` `= (dy)/(dx) xx (dx)/(dt)`
  `(dy)/(dx)` `=((dy)/(dt))/((dx)/(dt))`
  `:. (dy)/(dx)|_(t=1)` `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)`
    `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))`
    `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)`
    `= (1 + sqrt 3)/(1 – sqrt 3)`

 

d.   `text(At)\ \ t=0:` 

♦♦ Mean mark part (d) 32%.

`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`

`theta_1` `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4`
`theta_2` `= tan^(-1)(1/sqrt 3)=pi/6`

 
`text(Let)\ \ theta=\ text(angle between the vectors:)`

`theta` `= pi – theta_1 – theta_2`
  `= pi – pi/4 – pi/6`
  `= (12 pi – 3 pi – 2 pi)/12`
  `= (7 pi)/12`

Vectors, SPEC1 2013 VCAA 7

The position vector  `underset ~r (t)`  of a particle moving relative to an origin `O` at time `t` seconds is given by

`underset ~r(t) = 4 sec (t) underset ~i + 2 tan (t) underset ~j,\ t in [0, pi/2)`

where the components are measured in metres.

  1. Show that the cartesian equation of the path of the particle is  `x^2/16-y^2/4 = 1.`   (2 marks)

  2. Sketch the path of the particle on the axes below, labelling any asymptotes with their equations.   (2 marks)

     

        
     VCAA 2013 spec 7b
     

  3. Find the speed of the particle, in `text(ms)^-1`, when `t = pi/4.`   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
  3. `4 sqrt 3\ \ text(ms)^-1`
Show Worked Solution
a.   `x` `= 4sec(t)`
  `x/4` `= sec(t)`
  `y` `= 2tan(t)`
  `y/2` `= tan(t)`

  
`text(Using)\ \ tan^2 (t) + 1 = sec^2(t),`

`(x^2)/16 +1 ` `= y^2/4`
`:. (x^2)/16-(y^2)/4` `=1`

 

b.   `y^2/4 = x^2/16 -1\ \ =>\ \ y=+- sqrt(x^2/4 -4)`

♦ Mean mark part (b) 45%.

`lim_(x->oo) y = +- x/2`

 

♦ Mean mark part (c) 39%.

c.    `overset·underset~r(t)` `= d/(dt)(4(cos(t))^(−1))underset~i + d/(dt)(2tan(t)) underset~j`
    `= 4(−1)(−sin(t))(cos(t))^(−2)underset~i + 2sec^2(t)underset~j`
    `= 4sin(t)sec^2(t)underset~i + 2sec^2(t)underset~i`

 

`|overset·underset~r(pi/4)|` `= sqrt(16sin^2(pi/4)sec^4(pi/4) + 4sec^4(pi/4))`
  `= sqrt(16(1/sqrt2)^2(sqrt2)^4 + 4(sqrt2)^4)`
  `= sqrt(16(1/2)(4) + 4(4))`
  `= sqrt(48)`
  `= 4sqrt3\ \ text(ms)^(-1)`

Vectors, SPEC2 2018 VCAA 4

Two yachts, `A` and `B`, are competing in a race and their position vectors on a certain section of the race after time  `t`  hours are given by

`underset ~ r_A (t) = (t + 1) underset ~i + (t^2 + 2t) underset ~j \ and \ underset ~r_B (t) = t^2 underset ~i + (t^2 + 3) underset ~j, \ t >= 0`

where displacement components are measured in kilometres from a given reference buoy at origin `O`.

  1. Find the cartesian equation of the path for each yacht.   (2 marks)

  2. Show that the two yachts will not collide if they follow these paths.   (2 marks)

  3. Find the coordinates of the point where the paths of the two yachts cross. Give your coordinates correct to three decimal places.   (2 marks)

One of the rules for the race is that the yachts are not allowed to be within 0.2 km of each other. If this occurs there is a time penalty for the yacht that is travelling faster.

  1. For what values of `t` is yacht `A` travelling faster than yacht `B`?   (2 marks)

  2. If yacht `A` does not alter its course, for what period of time will yacht `A` be within 0.2 km of yacht `B`? Give your answer in minutes, correct to one decimal place.   (2 marks)

Show Answers Only
  1. `y_A = x_A^2-1, x_A > 1; qquad y_B = x_B + 3, x_B > 0`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `(x, y) ~~ (2.562, 5.562)`
  4. `0 < t < 5/2`
  5. `4.1\ text(minutes)`
Show Worked Solution

a.  `x_A = t + 1,\ \ y_A = = t^2 + 2t`

`y_A` `= t^2 + 2 t + 1-1`
  `= (t + 1)^2-1`
`:. y_A` `= x_A^2-1`

 
`x_B = t^2,`

`y_B` `= t^2+3`
`:.y_B` `= x_B+3`

 

b.  `text(Same)\ \ xtext(-coordinate occurs when:)`

`t+1` `= t^2`
`t` `= (1+sqrt5)/2,\ \ \ (t>0)`

 
`text(When)\ \ t = (1+sqrt5)/2,`

`y_A=(3sqrt5+5)/2`

`y_B= (sqrt5+9)/2  !=y_A`
 

`:.\ text(No collision)`

 

c.   `x + 3` `= x^2-1`
  `x^2-x-4` `= 0`
  `x` `= (1 + sqrt 17)/2,\ \ \ (t>0 \ =>\ x>1)`
  `y` `= (7 + sqrt 17)/2`
  `:. (x, y)` `= ((1 + sqrt 17)/2, (7 + sqrt 17)/2)`
    `=(2.562, 5.562)`

 

d.    `underset ~dot r_A (t)` `= underset ~i + (2t + 2)underset ~j`
  `|underset ~dot r_A (t)|` `= sqrt(1 + (2t + 2)^2)`
    `= sqrt(4t^2 + 8t + 5)`
     
  `underset ~dot r_B (t)` `= 2t underset ~i + 2t underset ~j`
  `|underset ~dot r_B (t)|` `= sqrt (4t^2 + 4t^2)`
    `= sqrt(8t^2)`

  
 `text(Yacht A is travelling faster when:)`

♦ Mean mark part (d) 41%.

`sqrt(4t^2 + 8t + 5)` `> sqrt (8t^2)`
`4t^2 + 8t + 5` `> 8t^2`
`4t^2-8t-5` `< 0`

 
`:. 0 < t < 5/2`

 

e.    `underset ~ r_B-underset ~r_A` `= (t^2-(t + 1)) underset ~ i + (t^2 + 3-(t^2 + 2t)) underset ~j`
    `= (t^2-t-1) underset ~i + (-2t + 3) underset ~j`

 

`d = |underset ~r_B-underset~r_A|= sqrt ((t^2-t-1)^2 + (3-2t)^2)`
 

`text(Find)\ \ t\ \ text(such that:)`

♦ Mean mark part (e) 37%.

`sqrt ((t^2-t-1)^2 + (3-2t)^2) < 0.2`

`=> 1.52883 < t < 1.59734`

 
`:.\ text(Period of time yachts are within 0.2 km)`

`~~ 1.59734-1.52883`

`~~0.068506\ text(hours)`

`~~ 4.1\ text(minutes)`

Vectors, SPEC1 2014 VCAA 2

The position vector of a particle at time  `t >= 0`  is given by

`underset ~r (t) = (t-2) underset ~ i + (t^2-4t + 1) underset ~j`

  1.  Show that the cartesian equation of the path followed by the particle is  `y = x^2-3`.  (1 mark)

  2.  Sketch the path followed by the particle on the axes below, labelling all important features.  (2 marks)


 

 

  1.  Find the speed of the particle when  `t = 1`.  (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `sqrt 5`

Show Worked Solution

a.    `x` `= t-2\ \ =>\ \ t=x+2`
`y` `= t^2-4t + 1`
  `= (x + 2)^2-4 (x + 2) + 1`
`y` `= x^2 + 4x + 4-4x-8 + 1`
  `= x^2-3`

 

b.   `t >= 0`

`x = t-2\ \ =>\ \ x >= -2`

`y(-2)` `= (-2)^2-3`
  `= 1`
`y(0)` `= -3`

 

`0` `= x^2-3`
`x^2` `= 3`
`x` `= +- sqrt 3`

 

c.    `underset ~ dot r (t)` `= underset ~ dot i + (2t-4) underset ~j`
  `underset ~ dot r (1)` `= underset ~ dot i + (2-4) underset ~j`
    `= underset ~ dot i-2 underset ~j`
  `|underset ~ dot r(1)|` `= sqrt (1^2 + (-2)^2)`
    `= sqrt 5`

Vectors, SPEC2 2017 VCAA 15 MC

A body has displacement of  `3underset~i + underset~j`  metres at a particular time. The body moves with constant velocity and two seconds later its displacement is  `−underset~i + 5underset~j`  metres.

The velocity, in ms−1, of the body is

  1. `2underset~i + 6underset~j`
  2. `−2underset~i + 2underset~j`
  3. `−4underset~i + 4underset~j`
  4. `4underset~i - 4underset~j`
  5. `underset~i + 3underset~j`
Show Answers Only

`B`

Show Worked Solution
`Deltaunderset~s` `= (−1 – 3)underset~i + (5 – 1)underset~j`
  `= −4underset~i + 4underset~j`
`underset~v` `= (Deltaunderset~s)/t`
  `= (−4underset~i + 4underset~j)/2`
  `= −2underset~i + 2underset~j`

 
`=> B`

Vectors, SPEC2 2018 VCAA 13 MC

The position vector of a particle that is moving along a curve at time `t` is given by  `underset ~r(t) = 3 cos (t) underset ~i + 4 sin (t) underset ~j, \ t >= 0`.

The first time when the speed of the particle is a minimum is

  1. `3`
  2. `pi/2`
  3. `tan^(-1) (4/3)`
  4. `(3 pi)/2`
  5. `9` 
Show Answers Only

`B`

Show Worked Solution
`underset ~dot r(t)` `= -3 sin(t) underset ~i + 4 cos (t) underset ~j`
`|underset ~dot r(t)|` `= sqrt (9 sin^2(t) + 16 cos^2(t))`
  `= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
  `= sqrt (9 + 7 cos^2(t))`

 
`text(Minimised for)\ cos(t) = 0`

`:.  t_1 = pi/2`

`=>  B`