Soccer player #1 is 7.31 m from the goal, as the figure shows. If she kicks the ball directly into the net, the ball has a displacement labeled . If, on the other hand, she first kicks it to player #2, who then kicks it into the net, the ball undergoes two successive displacements, y and x. What are the magnitudes of (a)x, and (b)y. Help?

Question

Soccer player #1 is 7.31 m from the goal, as the figure shows. If she kicks the ball directly into the net, the ball has a displacement labeled . If, on the other hand, she first kicks it to player #2, who then kicks it into the net, the ball undergoes two successive displacements, y and x. What are the magnitudes of (a)x, and (b)y. Help?

Answer 1

To find the magnitudes of the two successive displacements, x and y, we need to use vector addition.

In this scenario, player #1 kicks the ball directly into the net, which results in a displacement labeled d.

To find the magnitude of d, we can use the Pythagorean theorem since it forms a right triangle. The distance player #1 is from the goal, 7.31 m, represents the side adjacent to angle theta (θ), and d represents the hypotenuse.

Therefore, we can use the equation:
d^2 = (adjacent)^2 + (opposite)^2
d^2 = (7.31 m)^2 + (0 m)^2
d^2 = 7.31^2
d = √(7.31^2)

So, magnitude of displacement d is √(7.31^2) = 7.31 m.

Now, player #2 receives the ball from player #1 and kicks it into the net. This creates two successive displacements, y and x.

To find the magnitude of y, we can subtract the magnitude of displacement d from the distance player #2 is from the goal.

The distance from player #2 to the goal is labeled "x + y", which represents the hypotenuse of a right triangle with x as the adjacent side and y as the opposite side.

Using the Pythagorean theorem again, we have:
(x + y)^2 = x^2 + y^2
(x + y)^2 = (7.31 m)^2
(x + y) = √(7.31^2)

Now, to find the magnitude of y, we can substitute the value of displacement d and solve for y:
(√(7.31^2))^2 = x^2 + y^2
7.31^2 = x^2 + y^2
7.31^2 - x^2 = y^2
y = √(7.31^2 - x^2)

To find the magnitude of x, we can subtract the magnitude of displacement y from the distance player #2 is from the goal:
(x + y) - y = x
x + y - y = x
x = √(7.31^2) - √(7.31^2 - x^2)

These equations give you the magnitudes of displacement x and displacement y when player #1 passes the ball to player #2 before it reaches the goal. Just substitute the value of x into the equation for y to find their respective magnitudes.