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Vectors, SPEC2 2024 VCAA 15 MC

The position of a moving body is given by  \(\underset{\sim}{ r }(t)=\sin (t) \underset{\sim}{ i }+\cos (2 t) \underset{\sim}{ j }\), where \(t\) is measured in seconds, for  \(t \geq 0\).

The motion of the body can be described as moving along a parabolic path given by

  1. \(y=1-2 x^2\), starting at \((0,1)\), reversing direction at \((1,-1)\) and then again at \((-1,-1)\), then returning to \((0,1)\) after \(2 \pi\) seconds.
  2. \(y=1-x^2\), starting at \((1,0)\), reversing direction at \((-1,0)\), then returning to \((1,0)\) after \(2 \pi\) seconds.
  3. \(y=1-2 x^2\), starting at \((0,1)\), reversing direction at \((1,-1)\) and then again at \((-1,-1)\), then returning to \((0,1)\) after \(\pi\) seconds.
  4. \(y=1-x^2\), starting at \((1,0)\), reversing direction at \((-1,0)\), then returning to \((1,0)\) after \(\pi\) seconds.
Show Answers Only

\(A\)

Show Worked Solution
\(x\) \(=\sin (t) \Rightarrow \text{Period}=2 \pi\)  
\(y\) \(=\cos (2t) \Rightarrow \text{Period}=\pi\)  
  \(=1-2 \sin ^2(t)\)  
  \(=1-2 x^2 \quad \text{(Eliminate B and D)}\)  

 
\(\text {Motion starts at \((0,1)\) when  \(t=0\).}\)

\(\text{At  \(t=\dfrac{\pi}{2}\), position \((1,-1)\)}\)

\(\text{At  \(t=\dfrac{3\pi}{2}\), position \((-1,-1)\)}\)

\(\text{At  \(t=2 \pi\), returns to \((0,1)\) after a full period.}\)

\(\Rightarrow A\)

♦♦ Mean mark 36%.

Vectors, SPEC2 2020 VCAA 15 MC

Two forces, `underset~F_(text(A)) = 4 underset~i - 2 underset~j`  and  `underset~F_(text(B)) = 2 underset~i - 5 underset~j`, act on a particle of mass 3 kg. The particle is initially at rest at position  `underset~i + underset~j`. All force components are measured in newtons and displacements are measured in metres.

The cartesian equation of the path of the particle is

  1. `y = x/2`
  2. `y = x/2 - 1/2`
  3. `y = ((x + 1)^2)/2 + 1`
  4. `y = ((x - 1)^2)/1 + 1`
  5. `y = x/2 + 1/2`
Show Answers Only

`E`

Show Worked Solution

`text(Net Force)\ (underset~F) = 4underset~i – 2underset~j +  2underset~i + 5underset~j = 6underset~i + 3underset~j`

`underset~a = underset~F/m = 2underset~i + underset~j`

♦ Mean mark 38%.

`underset~v = int_0^t 2underset~i + underset~j\ dt = 2tunderset~i + tunderset~j`

`underset~r` `= int_0^t 2tunderset~i + tunderset~j\ dt + (underset~i + underset~j)`
  `= t^2 underset~i + (t^2)/2 underset~j + underset~i + underset~j`
  `= (t^2 + 1)underset~i + ((t^2)/2 + 1)underset~j`

 
`x = t^2 + 1 \ => \ t^2 = x – 1`

`y` `= (t^2)/2 + 1`
  `= ((x – 1))/2 + 1`
  `= x/2 + 1/2`

 
`=>E`

Vectors, SPEC2 2019 VCAA 15 MC

A particle is moving along the `x`-axis with velocity  `underset~v = u underset~i`, where  `u`  is a real constant.

At time  `t = 0`, a force acts on the particle, causing it to accelerate with acceleration  `underset~a = alpha underset~j`, where  `alpha`  is a negative real constant.

Which one of the following statements correctly describes the motion of the particle for  `t > 0`?

  1. The particle show down, stops momentarily and then begins to move in the opposite direction to its original motion.
  2. The particle continues to travel along the `x`-axis with decreasing speed.
  3. The particle travels parallel to the `y`-axis.
  4. The particle moves along a circular arc.
  5. The particle moves along a parabola.
Show Answers Only

`E`

Show Worked Solution

`underset~a = alpha underset~j`

`text(Once force acts on particle:)`

`underset~v = u underset~i + alphat underset~j,\ \ \ (text(let)\ c_1 = 0)`

`underset~r = utunderset~i + alpha/2 t^2 underset~j,\ \ \ (text(let)\ c_2 = 0)`

`underset~r\ \ text(describes a parabola.)`

`=>E`

Vectors, SPEC2 2014 VCAA 17 MC

The acceleration vector of a particle that starts from rest is given by

`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j - 20 e^(−2t) underset ~k`, where `t >= 0`.

The velocity vector of the particle, `underset ~v(t)`, is given by

  1. `−8 cos(2t) underset ~i - 40 sin(2t) underset ~j + 40e^(−2t) underset ~k`
  2. `2 cos(2t) underset ~i + 10 sin(2t) underset ~j + 10e^(−2t) underset ~k`
  3. `(8 - 8 cos(2t)) underset ~i - 40 sin(2t) underset ~j + (40e^(−2t) - 40) underset ~k`
  4. `(2 cos(2t) - 2) underset ~i + 10 sin(2t) underset ~j + (10e^(−2t) - 10) underset ~k`
  5. `(4 cos(2t) - 4) underset ~i + 20 sin(2t) underset ~j + (20 - 20e^(−2t)) underset ~k`
Show Answers Only

`D`

Show Worked Solution
`underset ~ dot v(t)` `= underset ~a(t)`
`underset ~v(t)` `= int underset ~a (t)\ dt`
  `= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j + (10e^(-2t) + c_2) underset ~k`

 
`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`

`0` `= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j + (10e^0 + c_2) underset ~k`
`0` `= (2 + c_0) underset ~i + c_1 underset ~j + (10 + c_2) underset ~k `

 
`=> c_0 = -2, \ \  c_1 = 0, \ \  c_2 = -10`
 

`:. underset ~v(t) = (2 cos (2t) – 2) underset ~i + 10 sin(2t) underset ~j + (10e^(-2t) – 10) underset ~k`

`=> D`

Vectors, SPEC1 2017 VCAA 9

A particle of mass 2 kg with initial velocity  `3underset~i + 2underset~j`  ms−1 experiences a constant force for 10 seconds.

The particle's velocity at the end of the 10-second period  `43underset~i-18underset~j`  ms−1 .

  1. Find the magnitude of the constant force in newtons.   (2 marks)

  2. Find the displacement of the particle from its initial position after 10 seconds.   (3 marks)

Show Answers Only
  1. `4sqrt5 N`
  2. `230underset~i-80underset~j`
Show Worked Solution
a.    `u_i` `= 3` `v_i` `= 43`
  `u_j` `= 2` `v_j` `= −18`
`43` `= 3 + 10a`
`40` `= 10a`
`a_i` `= 4`

♦ Mean mark (a) 38%.

`−18` `= 2 + 10a_j`
`−20` `= 10a_j`
`a_j` `= −2`

 

`underset~a` `= 4underset~i-2underset~j`
`|underset~a|` `= sqrt(4^2 + (−2)^2)`
  `= sqrt(20)`
  `= 2sqrt5`

 

`:.|F|` `= 2 xx 2sqrt5`
  `= 4sqrt5 N`

 

b.  `underset~a = 4underset~i-2underset~j`

`underset~v` `= int4underset~i-2underset~j\ dt`
  `= (4t + C_i)underset~i + (−2t + C_j)underset~j\ \ \ (C_i, C_j ∈ R)`

 
`underset~v(0) = C_i underset~i + C_j underset~j = 3underset~i + 2underset~j`

`=> C_i = 3, C_j = 2`
 

`underset~v(t) = (4t + 3)underset~i + (2-2t)underset~j`

`underset~x(t)` `= int(4t + 3)underset~i + (2-2t)underset~j\ dt`
  `= (((4t^2)/2 +3t) + b_i)underset~i + ((2t-(2t^2)/2) + b_j)underset~j \ \ \ (b_i, b_j ∈ R)`
  `= (2t^2 + 3t + b_i)underset~i + (2t-t^2 + b_j)underset~j`

 
`underset~x(0) = b_iunderset~i + b_junderset~j`

`underset~0 => b_i = b_j = 0`

`underset~x(t) = (2t^2 + 3t)underset~i + (2t-t^2)underset~j`

`underset~x(10)` `= (2(10)^2 + 3(10))underset~i + (2(10)-10^2)underset~j`
  `= 230underset~i-80underset~j`

Vectors, SPEC1 2017 VCAA 7

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

Find the length of the path that the particle travels along the curve from  `t = 0`  to  `t = pi/4`.  (4 marks)

Show Answers Only

`3/4`

Show Worked Solution

`x(t) = (cos(t))^3 `

♦ Mean mark 50%.

`xprime(t)= −3cos^2(t)sin(t)`

 
`y(t) = (sin(t))^3`

`yprime(t)= 3sin^2(t)cos(t)`

 

`l` `= int_0^(pi/4)sqrt((xprime(t))^2 + (yprime(t))^2)\ dt`
  `= int_0^(pi/4)sqrt(9cos^4(t)sin^2(t) + 9sin^2(t)cos^2(t))\ dt`
  `= int_0^(pi/4)3sqrt(sin^2(t)cos^2(t)(cos^2(t) + sin^2(t)))\ dt`
  `= int_0^(pi/4)3sqrt(sin^2(t)cos^2(t))\ dt`
  `= 3int_0^(pi/4)sin(t)cos(t)\ dt`
  `= 3/2 int_0^(pi/4)(2sin(t)cos(t))\ dt`
  `= 3/2int_0^(pi/4)sin(2t)\ dt`
  `= 3/2[−1/2 cos(2t)]_0^(pi/4)`
  `= −3/4(cos(pi/2) – cos(0))`
  `= 3/4`

Vectors, SPEC1 2018 VCAA 10

The position vector of a particle moving along  a curve at time `t` seconds is given by 

   `underset ~r (t) = t^3/3 underset ~i + (text{arcsin}(t) + t sqrt (1 - t^2)) underset ~j, \ 0 <= t <= 1`,

where distances are measured in metres.

The distance `d` metres that the particle travels along the curve in three-quarters of a second is given by

`d = int_0^(3/4) (at^2 + bt + c)\ dt`

Find `a, b` and `c`, where  `a, b, c in Z`.  (5 marks)

Show Answers Only

`a = -1,\ \ b = 0,\ \ c = 2`

Show Worked Solution
`x` `= t^3/3`
`x′ (t)` `=t^2`

♦♦ Mean mark 26%.

`y` `= sin^(-1)(t) + t(1 – t^2)^(1/2)`
`y′(t)` `= 1/sqrt(1 – t^2) + sqrt(1 – t^2) + t(1/2(-2t)(1 – t^2)^(-1/2))`
  `= 1/sqrt(1 – t^2) + sqrt (1 – t^2) – t^2/sqrt(1 – t^2)`
  `= (1 – t^2)/sqrt (1 – t^2) + sqrt (1 – t^2)`
  `= sqrt(1 – t^2) + sqrt(1 – t^2)`
  `= 2 sqrt (1 – t^2)`

 

`d` `= int_0^(3/4) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
  `= int_0^(3/4) sqrt(t^4 + 4(1 – t^2))\ dt`
  `= int_0^(3/4) sqrt(t^4 – 4t^2 + 4)\ dt`
  `= int_0^(3/4) sqrt((t^2 – 2)^2)\ dt`

 
`text(S)text(ince distance travelled along curve is positive,)`

`text(When)\ \ 0 <= t <= 1 => sqrt((t^2 – 2)^2) = 2 – t^2 >= 0`

`d = int_0^(3/4) – t^2 + 0*t + 2\ dt`

`a = -1,\ \ b = 0,\ \ c = 2`