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harkovskaia [24]

2 years ago

5

two soccer players, mary and jane, begin running from nearly the same point at the same time. mary runs in an easterly direction

at 4.43 m/s, while jane takes off in a direction 60.2o north of east at 5.07 m/s. how long is it before they are 24.7 m apart? what is the velocity of jane relative to mary? enter first the x-component and then the y-component. how far apart are they after 4.08 s?

1 answer:

goblinko [34]2 years ago

5 0

The time taken before the two soccer players are 24.7 m apart is 3 seconds.

Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.

The distance between Jane and Mary after 4.08 seconds is 33.58 m.

<h3>Resultant velocity of Mary and Jane</h3>

Vertical component of the velocity;

Jane: Vy1 = V sinθ

Vy1 = 5.07 m/s x sin(60.2)

Vy1 = 4.4 m/s

Horizontal components of the velocity;

Jane: Vx1 = V cos θ

Vx1 = 5.07 m/s x cos60.2

Vx1 = 2.52 m/s

mary: Vx2 = 4.43 m/s

total horizontal components of the velocity = 4.43 m/s + 2.52 m/s = 6.95 m/s

<h3>Resultant velocity</h3>

V = √Vy² + Vx²

V = √(4.4² + 6.95²)

V = 8.23 m/s

<h3>Time taken before they are 24.7 m apart</h3>

t = 24.7 m / 8.23 m/s

t = 3 seconds

Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.

<h3>Distance between Jane and Mary after 4.08 seconds</h3>

d = vt

d = 8.23 m/s x 4.08 s

d = 33.58 m

Learn more about relative velocity here: brainly.com/question/17228388

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