Projectiles, 2D Motion, displacement and average velocity

[copitlory8]

Homework Statement

A bird watcher meanders through the woods, walking 0.30 km due east, 0.65 km due south, and 2.25 km in a direction 37.0° north of west. The time required for this trip is 3.00 h. Determine the magnitude and direction (relative to due west) of the bird watcher's
(a) displacement and
1 km (magnitude)
2° (direction north of west)
(b) average velocity.
3 km/h (magnitude)
4° (direction north of west)
Use kilometers and hours for distance and time, respectively.

gravity: 9.8m/s^2

Homework Equations

Vf = Vi + A(D)T
(D)X = Vi(D)T + (A(D)T^2)/2
Vf^2 = Vi^2 + 2A(D)X

The Attempt at a Solution

I drew a diagram and i still don't know what to do.

 

[Andrew Mason]

This is not an acceleration problem. It is just a matter of simple vector addition.

What are the North-South and East-West components of the last leg of his trip. Add those to the first two displacements to get the final displacement. Then determine the angle and length of the final displacement.

To find the average velocity, use the final displacement divided by time (3 hours).

AM

 

[kerimek]

I would first clarify the given information and make sure all units are consistent. In this case, the given distances are in kilometers and the given time is in hours, so I would convert the distances to meters and the time to seconds to be consistent with the given value for gravity (9.8m/s^2).

Next, I would break down the bird watcher's motion into three separate parts: eastward motion, southward motion, and northwestward motion. For each part, I would use the equations for 2D motion to calculate the displacement and average velocity.

For the eastward motion, the displacement is 0.30 km or 300 m to the right (east) and the average velocity is 0.10 km/h or 0.028 m/s to the right.

For the southward motion, the displacement is 0.65 km or 650 m downwards (south) and the average velocity is 0.22 km/h or 0.061 m/s downwards.

For the northwestward motion, the displacement is 2.25 km or 2250 m at a 37.0° angle north of west and the average velocity is 0.75 km/h or 0.21 m/s at a 37.0° angle north of west.

To find the overall displacement, I would use the Pythagorean theorem to add the three displacements together: 300^2 + 650^2 + 2250^2 = 5300 m. The direction of this displacement can be found using trigonometry: tanθ = 650/300, so θ = 63.4°. This means the direction of the displacement is 63.4° south of east, or 26.6° west of south.

To find the overall average velocity, I would first find the total distance traveled by adding the three distances together: 0.30 km + 0.65 km + 2.25 km = 3.20 km. Then, I would divide this distance by the total time of 3.00 h to get an average velocity of 1.07 km/h or 0.30 m/s. The direction of this velocity can be found using trigonometry: tanθ = 0.30/1.07, so θ = 16.9°. This means the direction of the average