A fire ant, searching for hot sauce in a picnic area, goes through three displac
ID: 1585984 • Letter: A
Question
A fire ant, searching for hot sauce in a picnic area, goes through three displacements along level ground: vector d1 for 0.29 m southwest (that is, at 45° from directly south and from directly west), vector d2 for 0.57 m due east (that is, directly east), vector d3 for 0.60 m at 72° north of east (that is 72° toward the north from due east). Let the positive x direction be east and the positive y direction be north.
I have found the x and y components for d1, d2, and d3 but im am not sure how to do the last parts.
2. -14 points HRW10 3.XP.007 A fire ant, searching for hot sauce in a picnic area, goes through three displacements along east (that is 720 toward the north from due east). Let the positive x direction be east and What are the following components? (a) x component of d 0.205 (b) y component of di 0.205 (c) x component of d2 57 component of d (e) x component of d (D y component of da (d) y component of da 571 Give the following values for the ant's net displacement. (9) x component b) y component ) magnitude (counterclockwise from due @ast) Assume the ant retuns directly to the starting point. ) How far does it move? (1) In what direction? o(counterclockwise from due east) Additional Materials O Prefes atomatically sends sone dats to Moalla so that we can improve your experience.Explanation / Answer
the displacements in vector notation can be expressed as
d1 =0.29cos45(-i) +0.29sin45 (-j) =d1x(i) +d1y (j) = - 0.205(i) -0.205(j)
d2 = 0.57 (i) =d2x
d3 =0.60cos 72 (i) +0.60sin 72 (j) = d3x (i) +d3y(j) =0.1854(i) +0.5706(j)
net displacement =d= d1 +d2+d3 =(-0.205 +0.57 +0.1854) i +( -0.205 +0 +0.5704 ) j
=0.5504 i +0.3654 j =d x i +dy j
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