SOLUTION: Dora was travelling a = 6km, 30° N of E, b = 3km, 60° S of E, and then c = 10km, 15° N of E. At the end of this trek, she realized that in the last leg she should have travelled
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Question 1203884: Dora was travelling a = 6km, 30° N of E, b = 3km, 60° S of E, and then c = 10km, 15° N of E. At the end of this trek, she realized that in the last leg she should have travelled c' = 4km, 75° W of N instead of the third path she originally took. In component form, what path should Dora take to reach her supposed destination? What is the displacement from her initial position?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think ii have a solution.
hopefully it's a corect solution.
i have checked it over several times but it's easy to get twisted around so i'm keeping my fingers crossed that i did it right.
i did the calculations in excel because it's easier to keep track of them that way.
it's still a chore, but you at least have a record that you can go back and referencde and double check without having to do all the calculations over again each time.
the component form of vercotrs is the x and y component of the triangle formed.
when you draw a vector, the hypotenuce of the triangle formed is the distance traveled along the path of the bector.
the horizontal leg of the right triangle is the x component and the vertical leg of the right triangle is the y component.
the length of the x component is equal to the hypotenuse length * the cosine of the angle formed.
the length of the y component is equal to the hypoetn7use length * the sine of the angle formed.
the x component is assumed to extend horizontally from the vertex of the angle.
the vetical component is assumed to extended from the end of the horizontal line to the end of the hypotenuse.
the diagrams i created will show these relaationships.
the base side of the triangle formed needs to be on the horizontal axis.
i drew some sketches of the angles formed.
point O is the origin of the overall graph.
it is assumed to be at point (0,0).
the graph sketches are shown below.
O to A is from the origin to point A.
A to B is from point A to point B.
B to C is from point B to point C.
B to C' is from point B to point C'.
C to C' is from point C to point C'.
in order to locate the different points on the overall graph, it was necessary to calculate the horizonal and vertical distances from the origin to each point.
O to A is from the origin to point A.
O to B is from the origin to point B.
O to C is from the origin to point C.
O to C' is from the origin to point C'.
the vertical individual distances from each point are combined to form the overall distance from the origin to that point.
likewise, the horizontal distances from each point are combined to form the ocerall distance from the origin to that point.
a case in point is from O to C.
for the overall vertical distance, you get 3 - 2.598 + 2.588 = 2.99.
O to A is added because it is going up.
A to B is subtracted becqause it is going down.
B to C is added because it is going up.
for the overall horizontal distance, you get 5.196 + 1.5 + 9.659 = 16.355.
O to A is added because it is going to the right.
A to B is added because it is going to the right.
B to C is added because it is going to the right.
when you go from B to C', the sketch needed to be adjusted because the base side of the triangle formed needed to be on the horizontal axis and not the vertical axis.
doing that, preserved the convention that the x component is equal to the hypotenuse * the cosine of the angle, and the y component is equal to the hypotenuse * the sine of the angle.
the adjustment was from 75 degrees west of east converted to 15 degrees east of west.
that allowed the x component of the triangle formed to line up with the vertex of the angle.
this made use of the fact that sine(angle) = cosine(90 - angle).
75 degrees west of north is the same as 15 degrees north of west because 15 degrees + 75 degrees = 90 degrees.
for B to C', 15 degrees north of west was modelled rather than 75 degrees west of north.
if you need further clarification of this, let me know and i'll do my best to make it more understandable.
the graph skethces are shown below:
the calculations were performed in excel because it was easier to keep track of them tht way and reuse of calculations alread done was facilitated.
it sill took a lot of work and a lot of thinking, but the record of what was done was easier to display.
all numbers were rounded to 3 dcimal places for easer of viewing.
the calculations, however, were performed with the unrounded numbers that were stored, but not displayed.
here is what the spreadsheet looked like.
since i wanted to show the angle in degrees and excel worked with angle in radians, i need to convert from degrees to radians so excel could perform the trig functions on the angle.
that's why you see the angle in degrees and in radians.
the formula for converting degrees to radians is radians = degrees * pi / 180.
the formula for converting radians to degrees is degrees = radians * 180 / pi.
if there is anything about this spreadsheet that confuses you, let me know and i'll explain as best i can.
finally, i plotted the points on a graph to show the overall paths and the order in which they were taken.
the path lines are straiaght, but i gent them to avoid crossing out the locataion coordinates shown.
first she went to A, then to B, then to C, then to C'.
here's the graph.
i'll be available to answer any questions you might have.
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