Question

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.50 m/s relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.9 m/s and is moving in a direction of 32,9° east of north, relative to the ground.Part AWhat is the magnitude of the ball's velocity relative to Juan?Part BWhat is the direction of the ball's velocity relative to Juan?

Answer 1

a) The magnitud will be
`7.75(m)/(s)`.

b) The direction of the ball's velocity relative to Juan will be
`\phi = 25.44^\circ`.

Step-by-step explanation:

`\theta = 32.9^\circ`

Juan's velocity relative to the ground :
`v_(j,g)=(0 ; 7.5)(m)/(s)`

Ball's velocity relative to the ground :
`v_(b,g)=(12.9sin(\theta ) ; 12.9cos(\theta ))(m)/(s)`

We know that Ball's velocity relative to the ground is Ball's velocity relative to Juan plus Juan's velocity relative to the ground. This is a very important notion that even extends to more complex mathematical problems (differentiation).

`v_(b,g)=v_(b,j) + v_(j,g)` ⇒
`v_(b,j) =v_(b,g)-v_(j,g)`

⇒
`v_(b,j)= (12.9sin(\theta ) ; 12.9cos(\theta ))(m)/(s)  - (0 ; 7.5)(m)/(s) = [tex](7 ; 10.83)(m)/(s)  - (0 ; 7.5)(m)/(s)`[/tex]

∴
`v_(b,j)= (7 ; 3.33)(m)/(s)`

a) The magnitud will be
`\left | v_(b,j)  \right |=√(7^2 + 3.33^2) =7.75(m)/(s)`.

b) The direction of the ball's velocity relative to Juan will be:

`tan(\phi )=(3.33)/(7)` ⇒
`\phi =arctan((3.33)/(7))`

∴
`\phi = 25.44^\circ`.