Jim’s dog Sparky runs 50 m northeast to a tree, then 70 m west to a second tree,
and finally 20 m south to a third tree.
a. Draw a picture and establish a coordinate system.
b. Calculate Sparky’s net displacement in component form.
c. Calculate Sparky’s net displacement as a magnitude and an angle.
East = x = 50 cos 45 -70 = -34.6
North = y = 50 sin 45 - 20 = 15.4
magnitude = sqrt (34.6^2 + 15.4^2 )
angle north of east = tan^-1 (-15.4/34.6)
a. To draw a picture and establish a coordinate system, we can assume that Jim's starting position is the origin, and each direction represents a positive or negative axis.
Let's label the first tree as Tree 1, the second tree as Tree 2, and the third tree as Tree 3. We can use x-axis to represent the east-west direction and y-axis to represent the north-south direction.
Here's a rough illustration of the coordinate system and the dog's movements:
|
|
| Tree 3
|
|
|/-------------\
Tree 1 | | Tree 2
__________|______________|__________
|<--- 50m --->|<--- 70m --->
b. To calculate Sparky's net displacement in component form, we need to determine the change in x-coordinate (Δx) and the change in y-coordinate (Δy) based on the given distances and directions.
- Sparky moves 50 m northeast, which means he goes 50 m in the positive x-axis direction and 50 m in the positive y-axis direction.
- Sparky then moves 70 m west, which means he goes 70 m in the negative x-axis direction but does not change his y-coordinate.
- Finally, Sparky moves 20 m south, which means he goes 20 m in the negative y-axis direction but does not change his x-coordinate.
By calculating the changes in x and y, we can determine Sparky's net displacement in component form:
Δx = 50 m - 70 m = -20 m
Δy = 50 m - 20 m = 30 m
Therefore, Sparky's net displacement in component form is (-20 m, 30 m).
c. To calculate Sparky's net displacement as a magnitude and an angle, we can use the Pythagorean theorem and trigonometry formulas.
The magnitude of the displacement (D) can be calculated as follows:
D = √(Δx^2 + Δy^2)
D = √((-20 m)^2 + (30 m)^2)
D = √(400 m^2 + 900 m^2)
D = √(1300 m^2)
D ≈ 36.06 m
The angle (θ) between the positive x-axis and Sparky's displacement vector can be calculated as follows:
θ = tan^(-1)(Δy/Δx)
θ = tan^(-1)(30 m/-20 m)
θ ≈ -56.31° (counterclockwise from the positive x-axis)
Therefore, Sparky's net displacement is approximately 36.06 m at an angle of -56.31° from the positive x-axis.
a. Picture and coordinate system:
Y
|
|
|
|
|
|
|
-------------------------------------------------X
/ T1\
/-----------------------\
| |
| Sparky |
| \____________/ |
| |
| |
| _______________|
| |
| T2 T3 |
In the coordinate system, let the +X direction be east and the +Y direction be north.
b. To calculate Sparky's net displacement in component form, we can break down each movement into its X and Y components:
50 m northeast = 35.36 m east + 35.36 m north (using Pythagorean theorem)
70 m west = -70 m east (opposite direction of X)
20 m south = -20 m north (opposite direction of Y)
Therefore, the net displacement in component form is:
ΔX = 35.36 m east - 70 m east = -34.64 m east
ΔY = 35.36 m north - 20 m north = 15.36 m north
c. To calculate Sparky's net displacement as a magnitude and an angle, we can use the formula:
Magnitude of net displacement = √(ΔX² + ΔY²)
Angle = arctan(ΔY / ΔX)
Plugging in the values from part b:
Magnitude of net displacement = √((-34.64 m)² + (15.36 m)²) = √(1199.4496 m² + 236.1696 m²) = √1435.6192 m² = 37.89 m (rounded to two decimal places)
Angle = arctan((15.36 m) / (-34.64 m)) = -23.91° (rounded to two decimal places)
Therefore, Sparky's net displacement is approximately 37.89 m at an angle of -23.91° from the positive X-axis.