How Do I Solve These Physics Vector Problems?

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The discussion focuses on solving physics vector problems involving displacement and velocity. The first problem requires calculating the resultant vector from four displacement vectors, resulting in a magnitude of 13.4 km and an angle of 43.8 degrees relative to due west. The second problem involves finding the north and west components of an ostrich's velocity, yielding 13.172 m/s north and approximately 7.0 m/s west. The third question pertains to calculating the magnitudes of two displacements in a soccer scenario, while the fourth involves determining displacement components after following specific directions on a treasure hunt. The overall emphasis is on applying trigonometry to resolve vector components effectively.
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Can someone help me with this...

1.) Displacement vector A points due east and has a magnitude of 3.07 km. Displacement vector B points due north and has a magnitude of 7.56 km. Displacement vector C points due west and has a magnitude of 9.67 km. Displacement vector D points due south and has a magnitude of 4.4 km. Find (a) the magnitude of the resultant vector A + B + C + D and (b) its angle relative to due west.

2.) The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. An ostrich is running at a speed of 14.8 m/s in a direction of 63 degrees north of west. What is the magnitude of the ostrich's velocity component that is directed (a) due north and (b) due west?

3.)Soccer player #1 is 9 m from the goal, as the figure shows. If she kicks the ball directly into the net, the ball has a displacement labeled A. If, on the other hand, she first kicks it to player #2, who then kicks it into the net, the ball undergoes two successive displacements, Ay and Ax. What are the magnitudes of (a) Ax, and (b) Ay. here is the pic for it...
http://publish.hometown.aol.co.uk/hmlisa1/images/untitled.jpg

4.) You are on a treasure hunt and your map says "Walk due west for 14.9 paces, then walk 20 degrees north of west for 80.9 paces, and finally walk due north for 76.4 paces." What is the magnitude of the component of your displacement in the direction (a) due north and (b) due west?

Please help me with those 4 questions...it's due tomorrow and i h ave no idea what to di..i'ts my first time taking physics and my prof didn't teach us how to do it yet...i really appreciate it..thanks a lot guys
 
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These problems are simple applications of Trig. Draw the vector on a coordinate system, draw triangles showing the decomposition of the vector on the coordinate system. Use trig to find the components.
 


Sure, I would be happy to help you with these questions. It's great that you are seeking help and trying to understand the material. Let's go through each question one by one and see if we can figure out the answers together.

1.) For this question, we need to find the resultant vector of A + B + C + D. To do this, we can use the Pythagorean theorem and trigonometry. First, let's draw a diagram to visualize the situation:

[insert diagram here]

Now, we can see that the resultant vector will have a magnitude of the sum of all the individual vectors. So, the magnitude of the resultant vector will be:

Magnitude = √(3.07^2 + 7.56^2 + 9.67^2 + 4.4^2)

= √(9.4249 + 57.1536 + 93.5089 + 19.36)

= √179.4474

= 13.4 km

Next, we need to find the angle of the resultant vector relative to due west. To do this, we can use the inverse tangent function. So, the angle will be:

Angle = tan^-1(∑y/∑x)

= tan^-1((7.56 + 4.4)/(3.07 + 9.67))

= tan^-1(11.96/12.74)

= tan^-1(0.938)

= 43.8 degrees

So, the magnitude of the resultant vector is 13.4 km and its angle relative to due west is 43.8 degrees.

2.) In this question, we need to find the magnitude of the ostrich's velocity component that is directed due north and due west. To do this, we can use trigonometry and the given information about the speed and direction of the ostrich.

(a) For the velocity component directed due north, we can use the sine function:

Velocity component due north = Velocity x sin(angle)

= 14.8 x sin(63 degrees)

= 14.8 x 0.89

= 13.172 m/s

So, the magnitude of the velocity component directed due north is 13.172 m/s.

(b) For the velocity component directed due west, we can use the cosine function:

Velocity component due west = Velocity x cos(angle)

= 14.
 
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