How Do You Calculate Total Displacement in Multi-Directional Vector Problems?

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To calculate total displacement in multi-directional vector problems, it's essential to accurately break down each movement into its x and y components. In the car's journey, the calculated total displacement is approximately 89.12 km, but the direction calculation using arctan may be incorrect due to component misinterpretation. For the beetles, the vector addition must be carefully managed to ensure the second beetle's path aligns with the first's final position, with errors likely stemming from incorrect assumptions about vector components. The use of the Pythagorean theorem is crucial for finding the magnitude of the resultant vector. Accurate component breakdown and direction calculations are vital for solving these vector problems correctly.
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Homework Statement



1.)A car is driven east for a distance of 52 km, then north for 34 km, and then in a direction 35° east of north for 27 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

a. Find the magnitude
b. Find the direction (counter clockwise from east)

The Attempt at a Solution



So I got a. using the unit vector method which gave me 49.49 for the y component, and 74.12 for the x component. The answer for part a is 89.1237 km.

Now, for part b., I'm doing arctan(49.49/74.12) which is equal to 33.73, but it marks me wrong.

Any idea of what's happening?

Homework Statement



2.) Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.50 m due east, then 0.80 m at 22° north of due east. Beetle 2 also makes two runs and the first is 1.6 m at 41° east of due north.

(a) What must be the magnitude of its second run if it is to end up at the new location of beetle 1?

(b) In what direction must it run?

The Attempt at a Solution



For part a., I am assuming that Vector A+Vector B= 0.5 i + 0.8(cos(22) i + sin(22) j) and Vector C= 1.6(cos(41) i + sin(41) j). Vector D would be the magnitude of beetle 2's second run.
So, it'll something like Vector D=Vector A+Vector B-Vector C=(1.24 i +.3 j)-(1.2 i +1.05 j)= .04 i + .75 j

And by using the Pythagorean theorem: .04^2 + .75^2 = .751 and once again, it marks me wrong.

What am I doing wrong?
 
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Marioqwe said:


So I got a. using the unit vector method which gave me 49.49 for the y component, and 74.12 for the x component. The answer for part a is 89.1237 km.

Now, for part b., I'm doing arctan(49.49/74.12) which is equal to 33.73, but it marks me wrong.

Any idea of what's happening?


For part a., I am assuming that Vector A+Vector B= 0.5 i + 0.8(cos(22) i + sin(22) j) and Vector C= 1.6(cos(41) i + sin(41) j). Vector D would be the magnitude of beetle 2's second run.
So, it'll something like Vector D=Vector A+Vector B-Vector C=(1.24 i +.3 j)-(1.2 i +1.05 j)= .04 i + .75 j

And by using the Pythagorean theorem: .04^2 + .75^2 = .751 and once again, it marks me wrong.

What am I doing wrong?
For the first one, I think you made the wrong assumption. Same for the C vector in second.
Maybe this could help: http://id.mind.net/~zona/mstm/physics/mechanics/vectors/introduction/introductionVectors.html"
 
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