three children are throwing a frisbee in a park. the first child throws the frisbee 32m[W14S] to the second child who then throws it 15m[E62S] to the third child. what is the total displacement of the frisbee?

Answer 1

To find the total displacement of the frisbee, we need to find the resultant vector of the two individual displacements.

First, let's break down the given displacements into their horizontal (east-west) and vertical (north-south) components.

For the first child's throw:
- Horizontal component: 32m * cos(14°) = 31.2199m (rounded to 4 decimal places) to the west.
- Vertical component: 32m * sin(14°) = 8.0923m (rounded to 4 decimal places) to the south.

For the second child's throw:
- Horizontal component: 15m * cos(62°) = 6.4264m (rounded to 4 decimal places) to the east.
- Vertical component: 15m * sin(62°) = 12.9590m (rounded to 4 decimal places) to the south.

To find the total horizontal displacement, we add the horizontal components together:
31.2199m (first child's throw to the west) + 6.4264m (second child's throw to the east) = 37.6463m (rounded to 4 decimal places) to the west.

Similarly, to find the total vertical displacement, we add the vertical components together:
8.0923m (first child's throw to the south) + 12.9590m (second child's throw to the south) = 21.0513m (rounded to 4 decimal places) to the south.

Finally, we can find the magnitude and direction of the total displacement using Pythagorean theorem and trigonometry:
Magnitude: sqrt((37.6463m)^2 + (21.0513m)^2) = 42.8319m (rounded to 4 decimal places).
Direction: arctan(21.0513m / 37.6463m) ≈ 28.7296° south of west (rounded to 4 decimal places).

Therefore, the total displacement of the frisbee is approximately 42.8319m at an angle of 28.7296° south of west.