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The point A has position vector $egin{pmatrix} 8 \ -6 \ 5 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \\ \ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \ \\ \\ \ 6 \\ \ \\ \ \\ 125 \\ \\ \ \\ \\ \ \\ \\ \ \\ \\ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 \\ \ \ 10 \\ 10 \\10 \\-10 \\ 6 \\ -10 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ 10 \\ \ 5 -6 \ \\ \\ 5 \\ \ \ \ \ \ \ \ - \\ 0\\ -1\\ 4 \\ \ \ \ \ \ \ 4 \\ \ \ \ \ \ \ 5 \\ \ 4 \\ -0\\ \\ -6 \\ -8 \\ \\ \\ \ \ \ \ * 25 \\ \ 12 \ 1 -0 -0 \ 5 +10 +10 +5 \ \ 0 0 + 1 + 1 + 0$ The line $egin{pmatrix} x \ y \ z \\ \ \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ 0 \\ -6 \\ 7 \\ 5 \ 1\ \ \ \ \ \ 1 12 \ 40\ 4\ \ 2 7 - - - -7 -2 7 -100 \ \\ \ \ \ \ \ \ \ \ \ -15-3 \ 17 +6 +6 +60 \\ 20 -0 10 0\ 31 10 \ \\ \ 0 -30 \\ \ \ \ \ \ \ \ \ 10 \\ \ 5 \\ The point B lies on $ ext{l}$ and has position vector $egin{pmatrix} p \ 1 \ 2 p \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ -\ 0 \\ 2 \ \ \ \ \ \ 2 + 10 \ \ \ 7 \ \\ 2 \ \ \ +2 12 \ \ \ 5 \ \ \ + 8\ 0 \ \ \ \ \ \ \ 0 \ \ \ \ \ \10 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ 0 \ 6+12-25-8-0-0 \ 1 \ \ \ \ \ +12 - \ 2 \-0 \\ -0\ \ 0 \ \ \ \ 6 +40 -6 +6 -10 -0 -80 -100 \ 46 +5 \ \ \ \ \ +100 -10 8 0 +25 +20 \\ \ \ \ \ -0\ 4 +4 -4 -20 + 2 + 12 +12 - 10 + 10 +10 +40 +45 +30 +30 +20 +40 +8 +100$ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2024 - Paper 1
Question 13
The point A has position vector $egin{pmatrix} 8 \ -6 \ 5 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \... show full transcript
Worked Solution & Example Answer:The point A has position vector $egin{pmatrix} 8 \ -6 \ 5 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \\ \ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \ \\ \\ \ \\ \\ \ 6 \\ \ \\ \ \\ 125 \\ \\ \ \\ \\ \ \\ \\ \ \\ \\ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 \\ \ \ 10 \\ 10 \\10 \\-10 \\ 6 \\ -10 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ 10 \\ \ 5 -6 \ \\ \\ 5 \\ \ \ \ \ \ \ \ - \\ 0\\ -1\\ 4 \\ \ \ \ \ \ \ 4 \\ \ \ \ \ \ \ 5 \\ \ 4 \\ -0\\ \\ -6 \\ -8 \\ \\ \\ \ \ \ \ * 25 \\ \ 12 \ 1 -0 -0 \ 5 +10 +10 +5 \ \ 0 0 + 1 + 1 + 0$ The line $egin{pmatrix} x \ y \ z \\ \ \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ 0 \\ -6 \\ 7 \\ 5 \ 1\ \ \ \ \ \ 1 12 \ 40\ 4\ \ 2 7 - - - -7 -2 7 -100 \ \\ \ \ \ \ \ \ \ \ \ -15-3 \ 17 +6 +6 +60 \\ 20 -0 10 0\ 31 10 \ \\ \ 0 -30 \\ \ \ \ \ \ \ \ \ 10 \\ \ 5 \\ The point B lies on $ ext{l}$ and has position vector $egin{pmatrix} p \ 1 \ 2 p \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ -\ 0 \\ 2 \ \ \ \ \ \ 2 + 10 \ \ \ 7 \ \\ 2 \ \ \ +2 12 \ \ \ 5 \ \ \ + 8\ 0 \ \ \ \ \ \ \ 0 \ \ \ \ \ \10 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ 0 \ 6+12-25-8-0-0 \ 1 \ \ \ \ \ +12 - \ 2 \-0 \\ -0\ \ 0 \ \ \ \ 6 +40 -6 +6 -10 -0 -80 -100 \ 46 +5 \ \ \ \ \ +100 -10 8 0 +25 +20 \\ \ \ \ \ -0\ 4 +4 -4 -20 + 2 + 12 +12 - 10 + 10 +10 +40 +45 +30 +30 +20 +40 +8 +100$ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2024 - Paper 1
Step 1
Step 2
Hence, or otherwise, determine the shortest distance between the point A and the line $ ext{l}$.
Answer
To minimize the distance, we differentiate with respect to :
rac{d(|AB|^2)}{dp} = 12p - 24
Setting this to zero to find critical points:
Now substituting back into :
Thus, the shortest distance is:
Step 3
A particle is moving in simple harmonic motion, described by $x = -4(t + 1)$.
Answer
Given the position function, we can find the speed:
v = rac{dx}{dt} = -4
When the particle passes the origin, we set and solve for time:
The speed at this point is given as 4 m/s. The distance travelled in one complete cycle is:
Step 4
A particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to $v^2$, where $v$ ms$^{-1}$ is the speed of the particle, so that the acceleration is given by $a = -kv^2$.
Answer
Starting with:
rac{dv}{dt} = -kv^2 \\ ext{We can separate variables:} \\ rac{1}{v^2} dv = -k dt
Integrating:
rac{-1}{v} = kt + C \\ v = rac{1}{kt + C}
Substituting initial conditions to find C and k: v(0) = 40 ext{ gives us } 40 = rac{1}{C} \\ C = rac{1}{40} By substituting into the equation, we get: v = rac{40}{kt + rac{1}{40}}
To determine when the velocity is 30 m/s to the right:
Step 5