A hiker starts at his camp and moves the following distances while exploring his surroundings: 65.0 m north, 2.00 ✕ 102 m east, 1.10 ✕ 102 m at an angle 30.0° north of east, and 1.50 ✕ 102 m south.
(a) Find his resultant displacement from camp. (Take east as the positive x-direction and north as the positive y-direction.
Your answer made no sense idiot henry
herny is an idiot
Well, this hiker seems to be on quite the adventure! Let's help him out by finding his resultant displacement.
First, let's break down the distances the hiker moved:
- 65.0 m north
- 2.00 * 102 m east
- 1.10 * 102 m at an angle 30.0° north of east
- 1.50 * 102 m south
To simplify things, we can break down the distances into their x (east) and y (north) components:
- 65.0 m north = 0 m east + 65.0 m north
- 2.00 * 102 m east = 2.00 * 102 m east + 0 m north
- 1.10 * 102 m at an angle 30.0° north of east = 1.10 * 102 * cos(30.0°) m east + 1.10 * 102 * sin(30.0°) m north
- 1.50 * 102 m south = 0 m east - 1.50 * 102 m north
Now let's calculate the x and y components:
- X component: 2.00 * 102 m east + 1.10 * 102 * cos(30.0°) m east + 0 m east + 0 m east = 3.10 * 102 m east
- Y component: 65.0 m north + 1.10 * 102 * sin(30.0°) m north - 1.50 * 102 m north = 6.68 m north
Lastly, let's find the resultant displacement by using the Pythagorean theorem:
Resultant Displacement = √((3.10 * 102)^2 + (6.68)^2)
And once you've done the calculations, you'll have the hiker's resultant displacement from camp!
To find the resultant displacement from the hiker's camp, we need to add the individual displacement vectors together. The displacement vectors can be represented by their x and y components.
Given information:
- North displacement: 65.0 m
- East displacement: 2.00 ✕ 10^2 m
- Displacement at 30.0° north of east: 1.10 ✕ 10^2 m
- South displacement: 1.50 ✕ 10^2 m
To calculate the x and y components of each displacement vector, we can use trigonometry.
1. North displacement: The displacement is purely along the y-axis (up or positive direction). Therefore, the x-component is 0 m and the y-component is 65.0 m.
2. East displacement: The displacement is purely along the x-axis (right or positive direction). Therefore, the x-component is 2.00 ✕ 10^2 m and the y-component is 0 m.
3. Displacement at 30.0° north of east: To find the x and y components, we need to split the displacement into its horizontal and vertical components. Using trigonometry, we can determine:
- x-component = magnitude * cos(angle) = (1.10 ✕ 10^2 m) * cos(30.0°)
- y-component = magnitude * sin(angle) = (1.10 ✕ 10^2 m) * sin(30.0°)
4. South displacement: The displacement is purely along the y-axis (down or negative direction). Therefore, the x-component is 0 m and the y-component is -1.50 ✕ 10^2 m.
Now, let's calculate the x and y components for the displacement at a 30.0° angle north of east:
- x-component = (1.10 ✕ 10^2 m) * cos(30.0°) = (1.10 ✕ 10^2 m) * (0.866) ≈ 95.3 m
- y-component = (1.10 ✕ 10^2 m) * sin(30.0°) = (1.10 ✕ 10^2 m) * (0.5) = 55.0 m
Now, let's add all the x and y components together:
- x-component total = 2.00 ✕ 10^2 m + 95.3 m = 295.3 m (east direction is positive)
- y-component total = 65.0 m + 55.0 m - 1.50 ✕ 10^2 m = -30.0 m (north direction is positive, south direction is negative)
Finally, the resultant displacement is the vector formed by the x and y components:
Resultant displacement = √[(x-component total)^2 + (y-component total)^2]
Resultant displacement = √[(295.3 m)^2 + (-30.0 m)^2] ≈ 296.5 m
Therefore, the hiker's resultant displacement from the camp is approximately 296.5 m.
D=65m[90o] + 200[0o] + 110[30o] + 150[270].
X = 200 + 110*Cos30 = 295.3 m.
Y = 65 + 110*sin30 - 150 = -30 m., Q4.
Tan A = Y/X = -30/295.3 = -0.10159.
A = -5.80o = 5.8o S. of E. = Direction.
D = X/Cos A = 295.3/Cos5.8 = 297 m.