2. In the same tournament, a player is positioned 35 m (40° W of S] of the net. He shoots the puck 25 m [E] to a teammate. What second displacement does the puck have to travel in order to
make it to the net? Find the answer using the component method and one of the other two
methods discussed.

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!!!

answer: i think he can see forward ten seconds and move forward ten seconds but giorno can undo his actions with ger.

explanation:

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )