How Do You Calculate Resultant Displacement and Direction in Vector Problems?

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Homework Help Overview

The discussion revolves around calculating the resultant displacement and direction in a vector problem involving a car's movement. The problem involves a car traveling 20 km due north and then 35 km at an angle of 60 degrees west of north.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods for calculating the resultant displacement, including the use of the cosine law and sine law. Questions arise regarding the setup of the problem, the definition of displacement, and the appropriate formulas to use.

Discussion Status

There is an active exploration of various approaches to the problem, with some participants suggesting the use of right triangle methods while others advocate for the cosine law. Guidance has been offered regarding the relationships between the sides and angles of the triangle formed by the car's path.

Contextual Notes

Participants are considering the implications of the angles and sides in the context of vector addition and displacement, with some uncertainty about how to properly apply the formulas discussed.

r-soy
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Hi all

A car travels 20 km due north and then 35 km in a direction 60 degree west of north. Find the magnitude and direction of the car's resultant displacement.

How I can solve like this question ?

I try to solve :

the magnitude = R = A + B

= 55 km

direction of the car's resultant displacement

is north
 
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Do you understand what displacement is?

If I go 1km north and then 1km east, my final displacement is the distance (magnitude) and direction I am from my starting position. So the answer to this problem would be 45o East of North with a distance of \sqrt{2}km.
 
can I solve this question by this two rule for

magnitude = R = sqrt(A^2 + B^2)
dircaion = tan = y/x

?
 
Yes but you have to think a little more about the problem to set it up into a right-triangle so you can use those formulas.
 
I would not do it as a right triangle problem. You have a triangle with two sides of lengths 20 and 35. Since the first leg is "due north" and the second leg is "60 degrees" west of north", the angle between the two legs has measure 180- 60= 120 degrees. Since you have two sides of the triangle and the angle between them, use the "cosine law": c^2= a^2+ b^2- 2ab cos(C) where a and b are two sides of the triangle, C is the angle between them, and c is the third side, opposite C.

Here, c^2= 20^2+ 35^2- 2(20)(35)cos(120)
 
Moderator's note: this thread has been moved from General Math to the Homework & Coursework Questions area of the forum.
 
HallsofIvy said:
I would not do it as a right triangle problem. You have a triangle with two sides of lengths 20 and 35. Since the first leg is "due north" and the second leg is "60 degrees" west of north", the angle between the two legs has measure 180- 60= 120 degrees. Since you have two sides of the triangle and the angle between them, use the "cosine law": c^2= a^2+ b^2- 2ab cos(C) where a and b are two sides of the triangle, C is the angle between them, and c is the third side, opposite C.

Here, c^2= 20^2+ 35^2- 2(20)(35)cos(120)



thank you but c (third side ) mean magnitude ??

and now how to find the dircaion ??
 
Since you are given 2 sides of the triangle (and can calculate the third) and one angle, you could use the sine law to find the other angles: If A, B, and C are three angles in a triangle and a, b, and c are the lengths of the sides opposite each side, respectively, then
\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}
 

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