Solve Vector Car Problem: Find Mag & Dir of Displacement

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In summary: There are other ways to find the resultant, which may be easier for the given problem, but the law of cosines and law of sines will always work.In summary, the problem involves finding the magnitude and direction of a car's resultant displacement after traveling 20.0 km due north and 35.0 km in a direction 60 degrees northwest. The law of cosines is used to find the magnitude of the resultant, and the law of sines is used to find the direction. Alternatively, the resultant can also be found by breaking up the vectors into their x and y components and using Pythagoras and trigonometric functions.
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Violagirl
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Homework Statement



I was going to see if anyone might be able to provide with clarification on how to go about solving this problem? It was done in my book as an example but I didn't follow the logic to solving it.


A car travels 20.0 km due north and then 35.0 km in a direction 60 degrees northwest as shown in the attached document. Find the magnitude and direction of the car's resultant displacement.



Homework Equations



Relevant equations for vectors:

x = r cos Θ
y = r sin Θ
r = √x2 + y2
tan Θ = y/x

The Attempt at a Solution





In the problem, they use the law of cosines to find R (the magnitiude of the car's displacement) from the equation:

R2 = A2 + B2 - 2AB cos Θ

To find Θ, they take:

Θ = 180° - 60° = 120°

From there, they set up the equation as:

R = √AA2 + B2 - 2AB cos Θ

In substituting the numbers given in the problem, they find a final answer of 48.2 km. For the equation above, I understand that the pythagorean theorem is applied for part of it and that they take the equation for x as we are measuring for displacement in the x direction and using the vectors of r as shown in the document. My question for this part, why do they find -2AB for it? Is it because there are two vectors?

For the direction, they solve for R in the northerly direction with angle beta. But they set it up as:

sin β/β = sin Θ/R

I don't follow the setup for it. I'm not sure why it's all set up as a fraction for the equation? Any input would be appreciated!
 

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  • #2
Violagirl said:

Homework Statement



I was going to see if anyone might be able to provide with clarification on how to go about solving this problem? It was done in my book as an example but I didn't follow the logic to solving it.


A car travels 20.0 km due north and then 35.0 km in a direction 60 degrees northwest as shown in the attached document. Find the magnitude and direction of the car's resultant displacement.



Homework Equations



Relevant equations for vectors:

x = r cos Θ
y = r sin Θ
r = √x2 + y2
tan Θ = y/x

The Attempt at a Solution





In the problem, they use the law of cosines to find R (the magnitiude of the car's displacement) from the equation:

R2 = A2 + B2 - 2AB cos Θ

To find Θ, they take:

Θ = 180° - 60° = 120°

From there, they set up the equation as:

R = √AA2 + B2 - 2AB cos Θ

In substituting the numbers given in the problem, they find a final answer of 48.2 km. For the equation above, I understand that the pythagorean theorem is applied for part of it and that they take the equation for x as we are measuring for displacement in the x direction and using the vectors of r as shown in the document. My question for this part, why do they find -2AB for it? Is it because there are two vectors?

For the direction, they solve for R in the northerly direction with angle beta. But they set it up as:

sin β/β = sin Θ/R

I don't follow the setup for it. I'm not sure why it's all set up as a fraction for the equation? Any input would be appreciated!
There are several ways to calculate the resultant of two vectors, and using the law of cosines to find the magnitude of the resultant is one of them, which then requires using the law of sines to find the angle (direction) of the resultant (which you typed wrong , should be sin β/B = sin Θ/R).
It is often easier to find the resultant by breaking up each vector into its x and y components, and then algebraically add all the x components together to get R_x, add all the y components together to get R_y, and magnitude of R is the sq rt of the sum of those squares (per Pythagoras), and direction of R is the angle whose tangent is R_y/R_x. Always draw a rough sketch to approximate scale, and watch plus and minus signs.
 

1. What is a vector in relation to a car problem?

A vector is a mathematical representation of a quantity that has both magnitude (size or amount) and direction. In the context of a car problem, a vector can represent the displacement or movement of the car in terms of distance and direction.

2. How do I find the magnitude of displacement for a car?

To find the magnitude of displacement for a car, you can use the Pythagorean theorem. This involves squaring the horizontal and vertical components of the displacement and then taking the square root of the sum of these squared values. Alternatively, you can use trigonometric functions to calculate the magnitude.

3. How do I find the direction of displacement for a car?

The direction of displacement for a car can be found by using trigonometric functions. You can use the inverse tangent function to find the angle between the horizontal and vertical components of the displacement, which gives you the direction of the displacement.

4. What are some common units used for vector displacement?

Some common units used for vector displacement include meters (m), kilometers (km), and miles (mi) for distance, and degrees (°) or radians (rad) for direction. These units can vary depending on the specific problem and what is being measured.

5. How can I use vector displacement to solve a car problem?

Vector displacement can be used to solve a car problem by representing the car's movement as a vector and using mathematical operations to find the magnitude and direction of the displacement. This information can then be used to calculate other quantities such as speed, acceleration, and time. It is important to carefully consider the direction and units when using vector displacement in problem-solving.

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