Answer:
The displacement of the net from player 2 in component form = (-47.498î - 26.812j)
The displacement of the net from player 2 in statement form is 54.54 m and 29.44° (S of W) or 60.56° (W of S)
Explanation:
The sketch of the bearings described in the question is presented in the attached image to this solution.
Method 1
Using component method
Taking the player 1's position as the origin,
The displacement of the player 2 from the origin is (25î) m
The displacement of the net from the origin is 35[(sin θ)î + (cos θ)j]
But θ is the angle of the net's displacement reading from the positive x-axis in the anticlockwise direction. θ = 230°
Displacement of the net from the origin = 35[(cos 230°) + (cos 230°)]
= 35[-0.6428î - 0.7660j]
= (-22.498î - 26.812j) m
In component form, taking note of the directions of the respective displacements calculated (check the attached image)
(The displacement of the net from player 1) = (The displacement of player 2 from player 1) + (The displacement of the net from player 2)
Since we have agreed that player 1 is the origin
(The displacement of the net from origin) = (The displacement of player 2 from origin) + (The displacement of the net from player 2)
(-22.498î - 26.812j) = (25î) + (The displacement of the net from player 2)
The displacement of the net from player 2 = (-22.498î - 26.812j) - (25î) = (-47.498î - 26.812j)
The magnitude of this displacement = √[(-47.498)² + (-26.812)²]
= √(2,256.060004 + 718.883344) = 54.54 m
Direction = tan⁻¹ (-26.812/-47.498) = 209.44° (the signs on the components show that the direction is the third quadrant from the positive x-axis in the anti-clockwise direction)
Hence, the displacement of the net from player 2 is 54.54 m and 29.44° (S of W)
Method 2
Using trignometry,
We will use cosine and sine rule to obtain the required magnitude and direction of the displacement of the net from player 2
Cosine rule
Magnitude = √[35² + 25² - (2×25×35×cos 130°)] = √2,974.8783169514 = 54.54 m
Sine rule
(Sin θ)/35 = (Sin 130°)/54.54
Sin θ = (35 × Sin 130°)/54.54 = 0.4916
θ = Sin⁻¹ (0.4916) = 29.44°
This answer matches the answers from method 1.
Hope this Helps!!!
The second displacement that the puck will travel in order to make it to the net is 1.81 m.
The resultant displacement of the puck is calculated as follows;
A vertical line that intersects 35 (40° W of S) 25 m East, will become the adjacent side of the right triangle.
cos(50) = R/35
R = 35 x cos(50)
R = 22.5 m
Let the second displacement = x
Total horizontal displacement = 25 + x
Apply Pythagoras theorem to determine the second displacement.
(25 + x)² + 22.5² = 35²
(25 + x)² = 718.75
25 + x = √718.75
25 + x = 26.8
x = 26.8 - 25
x = 1.81 m
Thus, the second displacement that the puck will travel in order to make it to the net is 1.81 m.
Learn more about component vectors here: https://brainly.com/question/13416288
