1/3 points I Previous Answers SerPSE9 3 P067 A pirate has buried his treasure on

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1/3 points I Previous Answers SerPSE9 3 P067 A pirate has buried his treasure on an island with five trees located at the points (30.0 m, -20.0 m), (60.0 m, 80.0 m), (-10.0 m,-10.0 m), (40.0 m, -30.0 m), and (-70.0 m, 60.0 m), all measured relative to some origin, as shown in the figure below. His ship's log instructs you to start at tree A and move toward tree B, but to cover only one-half the distance between A and B. Then move toward tree C, covering one-third the distance between your current location and C. Next move toward tree D, covering one-fourth the distance between where you are and D. Finally move toward tree E, covering one-fifth the distance between you and E, stop, and dig My Notes Ask Your (a) Assume you have correctly determined the order in which the pirate labeled the trees as A, B, C, D, and E, as shown in the figure. What are the coordinates of the point where his treasure is buried? m, 10.668 Your response differs from the correct answer by more than 10%. Double check your calculations. m) (b) What if you do not really know the way the pirate labeled the trees? What would happen to the answer if you rearranged the order of the trees, for instance, to B (30 m, -20 m), A (60 m, 80 m), E (-10 m, -10 m), C (40 m, -30 m), and D (-70 m, 60 m)? State reasoning to show that the answer does not depend on the order in which the trees are labeled This answer has not been graded yet. Submit Answer Save Progress

Explanation / Answer

a)

As starting from point A,

Position vector of A,r1= rA = (30 i - 20 j) m

Positon vector of point B, rB = 60 i + 80 j

Positon vector of point ,C rC = -10 i - 10 j

Positon vector of point D, rD = 40 i - 30 j

Positon vector of point E, rE = -70 i + 60 j

displacement between A and B, rAB =   60 i + 80 j - (30 i - 20 j) = 30 i + 100 j

as you covers one half of rAB, so distance covered = 1/2 * (30 i + 100 j ) = 15 i + 50 j

so, distance covered to move r2, só position of r2  = (30 i - 20 j) + 15 i + 50 j = 45 i + 30 j

now, displacement of current position to rC , r2C= rC - r2 =  -10 i - 10 j - (45 i + 30 j) = -55 i -40 j

to move position r3 , you cover 1/3 of r2C ,

so, position of r3 = r2 + 1/3 * (-55 i -40 j) =  45 i + 30 j + 1/3 * (-55 i -40 j) = 26.67 i + 16.67 j

displacement of current position to rD , r3D = rD - r3 = 40 i - 30 j - (26.67 i + 16.67 j) = 13.33 i -46.67 j

to move position r4 , you cover 1/4 of r3D ,

so, position of r4 = r3 + 1/4 * r3D  =   26.67 i + 16.67 j + 1/4 *(13.33 i -46.67 j) = 30 i + 5 j

displacement of current position r4to rE , r4E = rE - r4 = -70 i + 60 j - (30 i + 5 j ) = -100 i + 55 j

to move new position r5 , you cover 1/5 of r4E ,

so, position of r5 = r4 + 1/4 * r4E = 30 i + 5 j + 1/5 *(-100i + 55 j) =  10 i + 16 j

só treasure is buried at position r5 = 10 i + 16 j

b) From the steps we have followed above we can see that

Ist position he moved from rA as r2=    rA + 1/2 * (rB - rA) = rA / 2 + rB / 2

from r2 moved to r3 as (rA / 2 + rB / 2 ) + (rc - (rA / 2 + rB / 2 ))/ 3 = rA /3 + rB /3 + rc / 3

from r3 moved to r4 as  rA /3 + rB /3 + rc / 3 + (rD - (rA /3 + rB /3 + rc / 3))/4 = rA / 4 + rB / 4 + rc / 4 + rD / 4

from r4 moved to r5 as rA / 4 + rB / 4 + rc / 4 + rD / 4 + (rE - (rA / 4 + rB / 4 + rc / 4 + rD / 4))/ 5

= rA / 5 + rB / 5 + rc / 5 + rD / 5 + rE / 5 = 1/5 * (rA + rB + rC + rD + rE)

from above result we can see that the position of treasure does not depend on the order in which tree are labeled so, treasure positionwill remain same for any order of labeled trees as 10 i + 16 j

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