Calculating Net Displacement Vector for Sparky's Run Using Law of Cosines

In summary, Jim's dog Sparky runs 50.0m northeast, 70.0m west, and 20.0m south to three different trees. To calculate Sparky's net displacement vector, an unknown triangle is drawn and the law of cosines is used to find an angle, but the value obtained is incorrect. The dog does not run in a triangle, so the picture is redrawn and it is determined that the simplest way to solve the problem is to reduce the first "run" to components. By doing so, it is found that Sparky runs 70(√(2)/2)= 35√(2) feet east and north, and the distances west and south are subtract
  • #1
mugzieee
77
0
Jim's dog Sparky runs 50.0m northeast to a tree, then 70.0m west to a second tree, and finally 20.0m south to a third tree.Calculate Sparky's net displacement vector.
I drew a picture and got an unknown triangle, which yileds the use of law of cos. I applies law of cos, to find an angle, and i get an angle to be 180 degrees. I know this value is wrong, but i know I will need at least one measure of an angle to compute the problem. I just don't see why the law of cos would go wrong here, can anybody help me out.
 
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  • #2
The dog does not run in a triangle, if he did, net displacement would be zero. Try redrawing the picture. I have the exact same physics text, keep in mind northeast implies exactly 45 degrees.
 
  • #3
Looks to me like the simplest way to do this problem is to reduce the first "run" to components. running 70 yards NE, Sparky runs 70(√(2)/2)= 35√(2) or about 49.5 feet east and 49.5 feet north. Now subtract the distances west and south from those to get the dogs final position. You can calculate distance and angle from his original position from that.
 
  • #4
or draw your diagram to scale. For example say 1 inch = 10 m.
 

1. What is physics-net displacement?

Physics-net displacement refers to the distance and direction an object travels from its starting point to its ending point. It takes into account the magnitude and direction of both linear and angular displacement.

2. How is physics-net displacement different from regular displacement?

Regular displacement only considers the distance an object travels in a straight line, while physics-net displacement takes into account both linear and angular displacement. This means that physics-net displacement is a more comprehensive measure of an object's overall movement.

3. What are some common units used to measure physics-net displacement?

The most common units used to measure physics-net displacement are meters (m) for linear displacement and radians (rad) for angular displacement. However, other units such as feet, kilometers, and degrees can also be used depending on the specific situation.

4. How is physics-net displacement related to velocity and acceleration?

Physics-net displacement is related to velocity and acceleration through the fundamental equations of motion. Velocity is the rate of change of displacement over time, while acceleration is the rate of change of velocity over time. Therefore, an object's physics-net displacement can be used to calculate its velocity and acceleration.

5. Can physics-net displacement be negative?

Yes, physics-net displacement can be negative. This means that the object has moved in a direction opposite to its initial direction. For example, if an object moves 5 meters to the right and then 3 meters to the left, its physics-net displacement would be 2 meters to the left, indicating a negative displacement.

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