Mario, a hockey player, is skating due south at a speed of 6.6 m/s relative to the ice. A teammate passes the puck to him. The puck has a speed of 12.9 m/s and is moving in a direction of 23° west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by Mario?

v(south) =12.9cos23 -6.6 =5.27 m/s

v(west) =12.9sin23=5.04 m/s
v=sqrt{v(west)²+v(south)²} =
=sqrt{5.04²+5.27²} =7.3 m/s
tanα =5.04/5.27 =0.95
α=43.7º

Why was the 6.6 subtracted because it's in the same direction as the 12.9cos23 vector?

Why are the south velocities subtracted because they are in the same direction?

Well, Mario certainly has his work cut out for him! Let's break down the situation, shall we?

The puck is already moving at a speed of 12.9 m/s in a direction 23° west of south, relative to the ice. Now, as Mario is skating due south, we need to find the velocity of the puck as observed by him.

To do that, we need to consider both the magnitude and direction. First, let's find the puck's southward component:

Puck's southward component = speed * sin(23°)
= 12.9 m/s * sin(23°)
≈ 5.11 m/s

Great! Now, let's find the puck's westward component:

Puck's westward component = speed * cos(23°)
= 12.9 m/s * cos(23°)
≈ 11.62 m/s

Alright, now we have the puck's southward and westward velocities. To find the magnitude, we can use the Pythagorean theorem:

Magnitude = √(southward velocity² + westward velocity²)
= √(5.11 m/s)² + (11.62 m/s)²
≈ √(26.12 + 135.05)
≈ √161.17
≈ 12.69 m/s

So the magnitude of the puck's velocity, as observed by Mario, is approximately 12.69 m/s.

Now, for the direction. Since Mario is skating due south, we can consider that his direction is 0° from south. The puck's direction would be the angle between its westward component and south, which we can determine using arctan:

Direction = arctan(westward velocity / southward velocity)
= arctan(11.62 m/s / 5.11 m/s)
≈ arctan(2.27)
≈ 66.94°

So, the direction of the puck's velocity, as observed by Mario, would be approximately 66.94° west of south.

Hope that helps clear things up! Now, let's see if Mario can score a goal with all this information. Good luck, Mario!

To find the magnitude and direction of the puck's velocity as observed by Mario, we need to break down the puck's velocity into its components.

1. First, let's find the north-south component of the puck's velocity. We can use the sine function since the angle is given relative to due south.

North-South Component = Puck's Velocity * sin(Angle)
= 12.9 m/s * sin(23°)
≈ 12.9 m/s * 0.3917
≈ 5.05 m/s (rounded to two decimal places)

2. Next, let's find the east-west component of the puck's velocity using the cosine function.

East-West Component = Puck's Velocity * cos(Angle)
= 12.9 m/s * cos(23°)
≈ 12.9 m/s * 0.9205
≈ 11.88 m/s (rounded to two decimal places)

Now that we have the north-south and east-west components, we can find the magnitude and direction of the puck's velocity as observed by Mario.

Magnitude = √(North-South Component)^2 + (East-West Component)^2
= √(5.05 m/s)^2 + (11.88 m/s)^2
≈ √(25.50 m²/s²) + (141.13 m²/s²)
≈ √166.63 m²/s²
≈ 12.91 m/s (rounded to two decimal places)

To find the direction relative to due south, we can use the trigonometric arctangent function.

Direction = arctan(North-South Component / East-West Component)
= arctan(5.05 m/s / 11.88 m/s)
≈ arctan(0.43)
≈ 23.37° (rounded to two decimal places)

Therefore, the magnitude of the puck's velocity as observed by Mario is approximately 12.91 m/s, and the direction is approximately 23.37° west of due south.