# Relative velocity passing a ball between two soccer players

ekpm
Homework Statement:
Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00 m/s. The velocity of the ball relative to Mia is 5.00 m/s in a direction 30.0 degrees east of south. What are the magnitude and direction of the velocity of the ball relative to the ground.
Relevant Equations:
V(B/G) = V(B/M) + V(M/G)
For some reason I'm having trouble understanding relative velocity problems. I know how to solve this, but I keep guessing at random methods until my answer matches the solution in the textbook.

I solved it correctly by breaking the velocity of the ball into x- and y- components, then solved for the components of V(B/G) (2.50 m/s and -4.33 m/s) and using the Pythagorean theorem to get 3.01 m/s and the angle arctan(1.67/2.50) = 33.7 degrees N of E.

But that's not the answer I got the first time when I drew a diagram. Why is the answer not V(B/G) = V(B/M) + V(M/G) = (-5.00 + 6.00) = 1.00 m/s in 30 degrees east of south (the same direction as the velocity of the ball relative to Mia.

I'm having trouble wrapping my head around what is going on. Alice has to be ahead of Mia kicking the ball back to Mia in order for it to travel east of south (Quadrant IV) in respect to Mia running north. But somehow the solution has the ball traveling North? This would mean the ball is traveling in two different directions at the same time because Mia is traveling North.

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The ball is traveling at 5 m/s relative to Mia, who is traveling at 6 m/s relative to the ground.

This means that even if the ball were going due south relative to Mia, it will be traveling north relative to the ground because 6 > 5.

ekpm
If the ball were going directly south relative to Mia with a magnitude of 5 m/s, would the vector for the V(B/M) be directly North? So the ball will kicked in front of her into the same direction she is running? Does the starting position of Alice even matter?

I read the subsection on relative velocities again and I can't wrap my head around this concept for some reason. I seem to miss every problem the first time around and rely on trial and error to finally get the answer, but still don't understand the concept. Like why can't I just add the two to get 1.00 m/s (or I guess 11.0 m/s if the vectors are in the same direction) and this would just be the magnitude).

EDIT: I might understand it now. If I'm running North at 6 m/s in the same direction of a ball that has a velocity of 0 m/s relative to me, the ball would be moving North at the same speed. If the ball is moving at 5 m/s North relative to me, it would be moving at 11 m/s relative to the ground. And if it is moving 5 m/s south relative to me it would be moving 1 m/s relative to ground?

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EDIT: I might understand it now. If I'm running North at 6 m/s in the same direction of a ball that has a velocity of 0 m/s relative to me, the ball would be moving North at the same speed.
It must be if you have the ball in your hand, for example!
If the ball is moving at 5 m/s North relative to me, it would be moving at 11 m/s relative to the ground. And if it is moving 5 m/s south relative to me it would be moving 1 m/s relative to ground?
Yes, exactly. These velocities are all in the same one dimension. In general, you need 2D or 3D vectors to represent motion and relative velocities on a 2D surface or in 3D space.

ekpm
It must be if you have the ball in your hand, for example!

Yes, exactly. These velocities are all in the same one dimension. In general, you need 2D or 3D vectors to represent motion and relative velocities on a 2D surface or in 3D space.
Thanks for confirming. I think two moving objects relative to each other threw me off. I just realized that if the ball is moving -6 m/s to me, it would be stationary because if I were running +6 m/s north.