Calculating Total Displacement of a Car and Finding Direction in Vector Homework

In summary, the car traveled 50 miles east, 30 miles north, and 25 miles in a direction 30 degrees east of north, starting from the origin. To find the total displacement, the three vectors must be added together, taking into account their angles. Using the Pythagorean theorem, the total displacement is found to be 81 miles from the starting point. The direction of the final displacement can be found by computing the east-west and north-south displacements separately and then using the Pythagorean theorem.
  • #1
courtrigrad
1,236
2
A car is driven eastward for a distance of 50 miles, then northward for 30 miles, and then in a direction 30 degrees east of north for 25 miles. Find the total displacement of the car from its starting point.

So I set up a vector diagram, starting at the origin. I did: 30^2 + 50^2 and took the square root of that to find the length of the first part. Then I added that to 25 and got 83.234, but the answer is 81.0. What am I doing wrong? Also how would you find the direction?

Thanks
 
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  • #2
What you are doing wrong is not recognizing that "displacement" is not the same as "distance traveled". If the car traveled 50 miles due east, then turned around and traveled 50 miles due west, it would have traveled 100 miles- but it certainly wouldn't be 50 miles from it's starting point, which is what "displacement" means!

Since you mention the car starts "at the origin", I presume that you want to use the vector components or "coordinates". Okay, setting up a coordinate system so that the unit vector i is East and unit vector j is west, the first vector is 50i (50 miles east), the second is 30j (30 miles north), and the third is 25 cos(30)i+ 25 sin(30)j.
Add those three vectors to find the final position of the car and use the Pythagorean theorem to find the distance that point is from the origin.
 
  • #3
i still get 83.3 while the answer in the back says 81.0. Originally I did [tex] 30^{2} + 50^{2} [/tex] and took [tex] \sqrt{3400} [/tex] which is 58.31. Then I added that to 25 and got 83.3! Also how would i finf the final direction?

thanks
 
  • #4
It's wrong to add the 25, as it's not in the same direction as the [tex]\sqrt{30^2+50^2}[/tex]

As HallsofIvy said:
[tex]\sqrt{(30+25cos(30))^2+(50+25sin(30))^2}[/tex]

Compute the total displacement in the east-west direction, and the total displacement in the north-south direction. Separately. Then add 'em together with the Pythagorean theorem.
 
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  • #5
The answer is correct is 81, the problem is you think the angle made by the first part is 60 degrees with the horizontal or 30 degrees with the vertical which is not the case, it's 30.96 degrees, that's why you find the slight difference between 81 and 83. Check Päällikkö's work for details.
 

1. How do you calculate total displacement of a car?

To calculate the total displacement of a car, you need to know the initial position and the final position of the car. Then, you can subtract the initial position from the final position to get the total displacement. This can be represented by the formula: Δx = xf - xi.

2. What is the difference between displacement and distance?

Displacement refers to the straight line distance between the initial and final positions, while distance takes into account the entire path traveled. Displacement is a vector quantity, while distance is a scalar quantity. In other words, displacement includes both magnitude and direction, while distance only includes magnitude.

3. How do you find the direction of a vector?

The direction of a vector can be found using trigonometric functions. First, you need to find the horizontal and vertical components of the vector, then use the inverse tangent function to find the angle between the vector and the horizontal axis. This angle represents the direction of the vector.

4. Can the total displacement of a car be negative?

Yes, the total displacement of a car can be negative. This indicates that the car has traveled in the opposite direction of its initial position. For example, if a car starts at position 5 and ends at position 2, the displacement would be -3 units.

5. How do you handle vectors that are not in a straight line?

When dealing with vectors that are not in a straight line, you can use the Pythagorean theorem to find the magnitude of the resulting vector. Then, you can use trigonometric functions to find the direction of the vector. Alternatively, you can break down the vector into its horizontal and vertical components and use vector addition to find the resulting vector.

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