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Vectors, SPEC2 2014 VCAA 19 MC

The velocity vector of a 5 kg mass moving in the cartesian plane is given by  `underset ~v(t) = 3 sin(2t) underset ~i + 4 cos(2t) underset ~j`, where velocity components are measured in m/s.

During its motion, the maximum magnitude of the net force, in newtons, acting on the mass is

  1. 8
  2. 30
  3. 40
  4. 50
  5. 70
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`C`

Show Worked Solution

`underset ~a(t) = underset ~dot v(t)= 6 cos(2t) underset ~i – 8 sin(2t) underset ~j`

`|\ underset ~a(t)\ |` `= sqrt(36 cos^2(2t) + 64 sin^2(2t))`
  `= sqrt(36 (cos^2(2t) + sin^2(2t)) + 28 sin^2(2t))`
  `= sqrt(36 + 28 sin^2 (2t))`

 

`|\ underset ~a(t)\ |_max` `= sqrt(36 + 28) qquad text(as max) (sin^2(2t)) = 1`
  `= 8`

 

♦ Mean mark 49%.

`:. |\ underset ~F\ |_max` `= 8 xx 5`
  `= 40`

`=> C`

Vectors, SPEC1 2016 VCAA 8

The position of a body with mass 3 kg from a fixed origin at time  `t`  seconds, `t >= 0`, is given by  `underset ~r = (3 sin (2t) - 2)underset ~i + (3 - 2 cos(2t)) underset ~j`, where components are in metres.

  1. Find an expression for the speed, in metres per second, of the body at time  `t`.  (2 marks)
  2. Find the speed of the body, in metres per second, when  `t = pi/12`.  (1 mark)
  3. Find the maximum magnitude of the net force acting on the body in newtons.  (3 marks)
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  1. `|underset ~ dot r| = sqrt(36 cos^2(2t) + 16 sin^2(2t))`
  2. `sqrt 31`
  3. `36`
Show Worked Solution
a.    `underset ~r ` `=(3 sin (2t) – 2)underset ~i + (3 – 2 cos(2t)) underset ~j`
   `underset ~ dot r` `= 6 cos (2t) underset ~i + 4 sin (2t) underset ~j`
  `|underset ~ dot r|` `= sqrt(36 cos^2(2t) + 16 sin^2(2t))`

 

b.   `text(Find)\ \ |underset ~ dot r|\ \ text(when)\ \ t=pi/12:`

`underset ~ dot r` `=(36 cos^2 (pi/6) + 16 sin^2 (pi/6))^(1/2)`  
  `= (36 (sqrt 3/2)^2 + 16 (1/2)^2)^(1/2)`  
  `= (36/4 xx 3 + 16/4)^(1/2)`  
  `= (27 + 4)^(1/2)`  
  `= sqrt 31`  

 

c.    `underset ~ ddot r` `= d/(dt) (underset ~dotr)= -12 sin (2t) underset ~i + 8 cos (2t) underset ~j`
  `underset ~F` `= 3 ddot r`
  `|underset ~F|` `= 3 |underset ~ ddot r|`
    `= 3 sqrt(144 sin^2 (2t) + 64 cos^2 (2t))`
    `= 3 sqrt(64 sin^2 (2t) + 64 cos^2(2t) + 80 sin^2 (2t))`
    `= 3 sqrt(64 + 80 sin^2 (2t))`

 
`text(Max occurs when)\ \ (sin^2 (2t)) = 1`

Mean mark 51%.

`:. max (|underset ~F|)` `= 3 sqrt(64 + 80)`
  `= 3 sqrt 144`
  `= 3 xx 12`
  `= 36`