smc-1179-75-Other applications

The position vector of the International Space Station `text{(S)}`, when visible above the horizon from a radar tracking location `text{(O)}` on the surface of Earth, is modelled by

`underset ~r(t) = 6800 sin(pi(1.3t - 0.1))underset~i + (6800 cos(pi(1.3t - 0.1)) - 6400)underset ~j`,

for `t in [0, 0.154]`,

where `underset ~i` is a unit vector relative to `text(O)` as shown and `underset ~j` is a unit vector vertically up from point `text(O)`. Time `t` is measured in hours and displacement components are measured in kilometres.

Find the height, `h` km, of the space station above the surface of Earth when it is at point `text(P)`, directly above point `text(O)`.

Give your answer correct to the nearest km (1 mark)
Find the acceleration of the space station, and show that its acceleration is perpendicular to its velocity. (3 marks)
Find the speed of the space station in km/h.

Give your answer correct to the nearest integer. (2 marks)
Find the equation of the path followed by the space station in cartesian form. (2 marks)
Find the times when the space station is at a distance of 1000 km from the radar tracking location `text(O)`.

Give your answers in hours, correct to two decimal places. (3 marks)

Show Answers Only

`text(400 km)`
`text(See Worked Solutions.)`
`27\ 772\ text(km/h)`
`x^2 + (y + 6400)^2 = 46\ 240\ 000`
`t ~~ 0.04, 0.11`

Show Worked Solution

a. `text(At)\ \ P,\ \ x(t)=0`

♦ Mean mark 49%.

`text{Solve (by CAS):}\ \ 6800 sin (pi(1.3t – 0.1))`

`= 0`

`=> t=0.076923…`

`:. h`	`= 6800 cos (pi (1.3(0.076923) – 0.1)) – 6400`
	`= 400`

b.	`underset ~(dot r) (t)`	`= 8840pi cos (pi(1.3t – 0.1)) underset ~i – 8840pi sin (pi(1.3t – 0.1)) underset ~j`
	`underset ~(ddot r) (t)`	`= -11\ 492 pi^2 sin (pi(1.3t – 0.1))underset ~i – 11\ 492 pi^2 cos (pi(1.3t – 0.1)) underset ~j`

`text(Let)\ \ u=pi(1.3t-0.1),`

`underset ~(ddot r) (t) ⋅ underset ~(dot r) (t)`

`= -(8840)(11\492) pi^3 cos (u) sin (u) + (8840)(11\492) pi^3 cos(u)sin(u)`

`= 0`

`:. underset ~(ddot r) (t) _|_ underset ~(dot r) (t)`

c. `text(Speed)\ = |\ dotr(t)\ |`

♦ Mean mark part (c) 45%.

`\|\ dotr(t)\ \|`	`= sqrt((8840pi)^2 cos^2(pi(1.3t – 0.1)) + (8840pi)^2 sin^2 (pi(1.3t – 0.1)))`
	`= 8840 pi`
	`~~ 27\ 772`

d.	`x/6800`	`= sin(pi(1.3t – 0.1))`
	`(6400 + y)/6800`	`= cos(pi (1.3t – 0.1))`

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

`x^2/6800^2 + (6400 + y)^2/6800^2`	`=1`
`x^2 + (y + 6400)^2`	`= 6800^2`

e. `underset ~r(t) = 6800 sin(u)underset~i + (6800 cos(u) – 6400)underset ~j`

♦ Mean mark part (e) 37%.

`text(Find)\ \ t\ \ text(such that)\ \ |\ underset ~r(t)\ | = 1000:`

`6800^2sin^2(u) + (6800 cos(u) -6400)^2 = 1000^2,\ \ \ (u=pi(1.3t-0.1))`

`6800^2 – 2 xx 6400 xx 6800 cos(u) + 6400^2 = 1000^2`

`=>cos(pi(1.3t-0.1)) = 0.9903\ \ \ text{(by CAS)}`

`:. t=0.04 or 0.11`

On a particular morning, the position vectors of a boat and a jet ski on a lake `t` minutes after they have started moving are given by `underset~r_B(t) = (1 - 2cos(t)) underset~i + (3 + sin(t))underset~j` and `underset~r_J(t) = (1 - sin(t)) underset~i + (2 - cos(t))underset~j` respectively for `t >= 0`, where distances are measured in kilometres. The boat and the jet ski start moving at the same time. The graphs of their paths are shown below.

On the diagram above, mark the initial positions of the boat and the jet ski, clearly identifying each of them. Use arrows to show the directions in which they move. (2 marks)
1. Find the first time for `t > 0` when the speeds of the boat and the jet ski are the same. (2 marks)
2. State the coordinates of the boat at this time. (1 mark)
1. Write down an expression for the distance between the jet ski and the boat at any time `t`. (1 mark)
2. Find the minimum distance separating the boat and the jet ski. Give your answer in kilometres, correct to two decimal places. (1 mark)
On another morning, the boat’s position vector remained the same but the jet skier considered starting from a different location with a new position vector given by `underset~r(t) = (1 - 2sin(t)) underset~i + (a - cos(t))underset~j, \ t >= 0`, where `a` is a real constant. Both vessels are to start at the same time.
Assuming the vessels would collide shortly after starting, find the time of the collision and the value of `a`. (3 marks)

Show Answers Only

1. `t = 0, t_2 = pi`
2. `(3,3)`
1. `sqrt((sin(t) – 2cos(t))^2 + (1 + sin(t) + cos (t))^2)`
2. `d_text(min) ~~ 0.33`
`a = 3 + 3/sqrt5, t = tan^(−1)(2)`

Show Worked Solution

`text(Initial positions occur at)\ \ t=0.`

`text(Consider the graph)\ \ underset~r_B(t)\ \ text(at)\ \ t=pi/4:`

`(1 – 2cos0) < (1 – 2cos(pi/4)) =>\ text(moves higher)`

`text(Consider the graph)\ \ underset~r_J(t)\ \ text(at)\ \ t=pi/4:`

`(1 – sin(0)) > (1 – sin(pi/4)) =>\ text(moves left)`

b.i.	`underset~dotr_B(t)`	`= 2sin(t)underset~i + cos(t)underset~j`
	`\|underset~dotr_B(t)\|`	`= sqrt(4sin^2(t) + cos^2(t))`

`underset~dotr_J(t)`	`= −cos(t)underset~i + sin(t)underset~j`
`\|underset~dotr_J(t)\|`	`= sqrt(cos^2(t) + sin(t))=1`

`text(Find)\ \ t\ \ text(when)\ \ sqrt(4sin^2(t) + cos^2(t))=1:`

`t = pi\ text(seconds)\ \ \ (t!=0)`

♦ Mean mark (b)(ii) 48%.

b.ii.	`(underset~r)_B(pi)`	`= (1 – 2cos(pi))underset~i + (3 + sin(pi))underset~j`
		`= 3underset~i + 3underset~j`

`:.\ text(Boat coordinates):\ (3,3)`

c.i.	`underset~r_B – underset~r_J`	`=(sin(t) – 2 cos(t))i +(1+sin(t) + cos(t))`
	`:. d`	`= \|underset~r_B – underset~r_J\|`
		`= sqrt((sin(t) – 2cos(t))^2 + (1 + sin(t) + cos (t))^2)`

c.ii. `d_text(min) ~~ 0.33\ \ \ text{(by CAS)}`

♦♦♦ Mean mark part (c)(ii) 15%.

d. `text(Equating coefficients for collision:)`

	`x:\ \ \ 1 – sin(t)`	`= 1 – 2cos(t)\ …\ (1)`
	`sin(t)`	`= 2cos(t)`
	`tan(t)`	`= 2`
	`t`	`=tan^(−1)(2)`

`y:\ \ \ a – cos(t) = 3 + sin(t)\ …\ (2)`

♦ Mean mark part (d) 41%.

`text(Using)\ \ tan(t)=2,`

`=> sin(t) = (2sqrt5)/5,\ \ cos(t) = sqrt5/5`

`text{Substitute into (1):}`

`a – 1/sqrt5`	`= 3 + 2/sqrt5`
`:. a`	`= 3 + 3/sqrt5`

`:.\ text(Collision occurs when)\ \ t=tan^(-1)2\ \ text(and)\ \ a=3 + 3/sqrt5`

Vectors, SPEC2 2012 VCAA 4

Vectors, SPEC2 2017 VCAA 5