Magnitude and Direction Word Problem

In summary, the pilot flew 45 miles west to another airport, then flew 20 miles at 15° east of south before finally landing at the third airport. The next day, he needs to fly directly to his original destination, and he has flown an extra 45 miles beyond his original flight plan. To find the x and y components for the second leg, drawing a diagram of the whole trip can be helpful. For the first leg, the x component is -45 miles and the y component is 0 miles. The correct way to find the resultant is by using the Pythagorean theorem and taking the square root of (-45)^2 + 5.17^2, which equals approximately 45.3 miles.
  • #1
slu1986
36
0
1. The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 45 miles west to another airport to find that conditions there are too icy for him to land. He flies 20 miles at 15° east of south and is finally able to land at the third airport.

(a) How far and in what direction must he fly the next day to go directly to his original destination?

(b) How many extra miles beyond his original flight plan has he flown?


3. I know that you have to solve get the x and y components and plug them into the pythagorean theorem to get the resultant. And then do the arc tangent y/x to get the direction in degrees. I am just confused at how to set everything up. I figure that the x component of the 1st position is -45 mi since he is going west. and y - component is the sin (15) 20 mi = 5.17 mi. and plug these numbers into the pythagorean theorum = sqrt of -45^2 + 5.17^2 = 45.3 mi <---but i know this isn't right..So can someone please help me out and explain what I'm doing wrong and what I must do to do it correctly
 
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  • #2
For the second leg, 20 miles at 15o East of South the vector has both x and y components. What are they? Drawing a diagram of the whole trip should help.
 
  • #3
Take another look at the first leg, especially the [tex]y[/tex]-component. As I read the problem, the first leg is due west.
 

FAQ: Magnitude and Direction Word Problem

1. What is a magnitude and direction word problem?

A magnitude and direction word problem is a type of problem that involves calculating the magnitude and direction of a vector or quantity. This means determining the size or amount of the vector, as well as the direction in which it is moving or acting.

2. How do you solve a magnitude and direction word problem?

To solve a magnitude and direction word problem, you first need to identify the given information, including the vector's magnitude and direction. Then, you can use trigonometric functions, such as sine, cosine, and tangent, to calculate the missing values. Finally, check your answer to make sure it makes sense in the context of the problem.

3. What are some real-life examples of magnitude and direction word problems?

Examples of magnitude and direction word problems in real life include calculating the velocity and direction of a moving object, determining the force and direction of a push or pull, and finding the magnitude and direction of a current in a river or an airplane's wind speed and direction.

4. Why is it important to understand magnitude and direction in science?

Understanding magnitude and direction is crucial in science because many physical quantities, such as velocity, force, and acceleration, are vectors and have both magnitude and direction. Without being able to accurately calculate and interpret these values, it would be challenging to make accurate predictions and understand the behavior of objects and systems.

5. What are some tips for solving magnitude and direction word problems?

Some tips for solving magnitude and direction word problems include drawing a diagram to visualize the problem, using trigonometric functions to find the missing values, and double-checking your answer to make sure it makes sense in the context of the problem. It can also be helpful to practice with different types of problems to become more familiar with the concepts and techniques involved.

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