In what direction is the scout's position relative to the campsite?

[neutron star]

Okay, the first one I have no clue how to set up or start, so a hunt would be nice. The second one, I figured out (a) but can''t seem to get (b) or (c) and want to know what I'm doing wrong.

Homework Statement

A scout walks 5 km due east from camp, then turns left and walks 4.5 km along the arc of a circle centered at the campsite, and finally walks 3 km directly toward the camp.
(a) How far is the scout from camp at the end of his walk?
___ km

(b) In what direction is the scout's position relative to the campsite?
___ ° (clockwise from north)

(c) What is the ratio of the final magnitude of the displacement to the total distance walked?
___





A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 3.0 m, y = 6.0 m, and has velocity v = 1.0 m/s i + 7.0 m/s j. The acceleration is given by the vector a = 9.0 m/s i + -1 m/s j.


What is the velocity vector at t=3.0s
I got (28 m/s) i+ (4 m/s) j

What is the position vector at t=5.0s

What is the magnitude and direction of the position vector

Homework Equations

The Attempt at a Solution

 

[Delphi51]

I would start with a diagram!
For the circle part, you know the distance from camp will be the same everywhere on the circle so that part is easy. You can get the angle through figuring out what part of the full circumference is walked, and take that same part of 360 degrees.

 

[tomcue]

I can provide a response to the given content by using mathematical principles and equations to analyze and interpret the given information. In the first problem, the scout's position can be described using vector addition. The scout initially walks 5 km due east, which can be represented as a vector of magnitude 5 km in the east direction. Then, the scout turns left and walks 4.5 km along the arc of a circle centered at the campsite, which can be represented as a vector with magnitude 4.5 km and direction perpendicular to the initial east vector. Finally, the scout walks 3 km directly toward the camp, which can be represented as a vector with magnitude 3 km in the west direction. To find the total displacement vector, we can add these three vectors together using the vector addition law. The resulting vector will be the scout's final position relative to the campsite.

(a) To find the magnitude of the final displacement vector, we can use the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this case, the final displacement vector can be represented as (5 km + 4.5 km + (-3 km))i, which simplifies to 6.5 km i. Therefore, the magnitude of the final displacement vector is 6.5 km.

(b) To find the direction of the final displacement vector relative to the campsite, we can use the inverse tangent function to find the angle between the vector and the positive x-axis. The resulting angle will be in the counter-clockwise direction from the positive x-axis. Therefore, the direction of the scout's position relative to the campsite is 49.5° clockwise from north.

(c) To find the ratio of the final magnitude of the displacement to the total distance walked, we can use the definition of displacement and distance. Displacement is the shortest distance between the initial and final positions, while distance is the total length of the path traveled. In this case, the total distance traveled is 5 km + 4.5 km + 3 km = 12.5 km, while the final displacement is 6.5 km. Therefore, the ratio of the final magnitude of the displacement to the total distance walked is 6.5/12.5 = 0.52.

In the second problem, we are given the position, velocity,