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Vectors, SPEC2-NHT 2019 VCAA 14 MC

The position vector  `underset~r(t)`  of a mass of 3 kg after `t` seconds, where  `t >= 0`, is given by  `underset~r(t) = 10tunderset~i + (16t^2 - 4/3t^3)underset~j`.

The force, in newtons, acting on the mass when  `t = 2`  seconds is

  1.  `16underset~j`
  2.  `32underset~j`
  3.  `48underset~j`
  4.  `30underset~i + 144underset~j`
  5.  `16`
Show Answers Only

`C`

Show Worked Solution

`underset~r(t) = 10tunderset~i + (16t^2 – 4/3t^3)underset~j`

`underset~v(t) = 10underset~i + (32t – 4t^2)underset~j`

`underset~a(t) = (32 – 8t)underset~j`

`text(When)\ t = 2,`

`underset~F` `= m underset~a`
  `= 3(32 – 16)underset~j`
  `= 48underset~j`

`=>\ C`

Vectors, SPEC2 2011 VCAA 11 MC

Consider the three forces

`underset~F_1 = −sqrt3underset~j`,  `underset~F_2 = underset~i + sqrt3underset~j`  and  `underset~F_3 = −1/2underset~i + sqrt3/2underset~j`.

The magnitude of the sum of these three forces is equal to

  1. the magnitude of `(underset~F_3 - underset~F_1)`
  2. the magnitude of `(underset~F_2 - underset~F_1)`
  3. the magnitude of `underset~F_1`
  4. the magnitude of `underset~F_2`
  5. the magnitude of `underset~F_3`
Show Answers Only

`E`

Show Worked Solution
`underset~(F_1) + underset~(F_2) + underset~(F_3)` `= (1 – 1/2)underset~i + (−sqrt3 + sqrt3 + sqrt3/2)underset~j`
  `= 1/2 underset~i + sqrt3/2underset~j`
`|underset~(F_1) + underset~(F_2) + underset~(F_3)|` `= sqrt((1/2)^2 + (sqrt3/2)^2)`
  `= |underset~(F_3)|`

`=> E`

Vectors, SPEC2 2016 VCAA 13 MC

A particle of mass 5 kg is subject to forces  `12 underset ~i`  newtons and  `9 underset ~j`  newtons.

If no other forces act on the particle, the magnitude of the particle’s acceleration, in ms¯², is

  1. `3`
  2. `2.4 underset ~i + 1.8 underset ~j`
  3. `4.2`
  4. `9`
  5. `60 underset ~i + 45 underset ~j`
Show Answers Only

`A`

Show Worked Solution
`underset ~F` `= 12 underset ~i + 9 underset ~j`
`5 underset ~a` `= 12 underset ~i + 9 underset ~j`
`underset ~a` `= 12/5 underset ~i + 9/5 underset ~j`
   
`|underset ~a|` `= sqrt(144/25 + 81/25)`
  `= 3`

 
`=>  A`

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 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)`
  `= sqrt(4 xx 5)`
  `= 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`