Question

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/s. The velocity of the ball relative to Mia is 3.60 m/s in a direction 30.0degree east of south.

A) What is the magnitude of the velocity of the ball relative to the ground?
Express your answer to three significant figures and include the appropriate units.

B)
What is the direction of the velocity of the ball relative to the ground ( North of East) ?
Express your answer to three significant figures and include the appropriate units.

Answer 1

a) v_{p} = 2.83 m / s , b) 50.5º north east

Step-by-step explanation:

This is a vector problem.

`v_(bg) = v_(bm) + v_(mg)`

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground

To make the sum we decompose the speed of the ball in its components

The angle of 30 east of the south, measured from the positive side of the x axis is

θ = 30 + 270 = 300

`v_(bx)` =
`v_(b)` cos 300

`v_(by)` = v_{b} sin. 300

v_{bx} = 3.60 cos 300 = 1.8 m / s

v_{by} = 3.60 sin 300 = -3,118 m / s

Let's add speeds on each axis

X axis

vₓ = v_{bx}

vₓ = 1.8 m / s

Y Axis

`v_(y)` = v1 - vpy

v_{y} = 5.30 - 3.118

v_{y} = 2.182 m / s

The magnitude of the velocity can be found using the Pythagorean theorem

`v_(p)` = √ (vₓ² + v_{y}²)

v_{p} = √ (1.8² + 2.182²)

v_{p} = 2,829 m / s

v_{p} = 2.83 m / s

b) for direction use trigonometry

tan θ =
`v_(y)` / vₓ

θ = tan ⁺¹ v_{y} / vₓ

θ = tan⁻¹ 2.182 / 1.8

Tea = 50.48º

This address is 50.5º north east