Question

Person A and person B are both riding bicycles, traveling due north at constant speeds. Person A is going at a velocity of 2.00 m/s and person B at a velocity of 8.00 m/s. Their paths are parallel and seperated by 1.5 m. At a given instant when B is 2.0 m behind A, B throws a rock due west relative to herself. The speed of the rock is such that A catched it without having to reach for it. Find the horizontal speed in m/s of the rock relative to Earth.

Answer 1

The main thing to notice in this problem is that if the rock's velocity is split up into "x and y" (even though neither velocity is vertical) the time it takes for the rock to cross the middle distance of l (l=1.5 meters) is the same amount of time that it takes for the rock to travel the Northernly distance, D. Setting up that equation:

The time it takes for the rock to travel 1.5 m horizontally:
\[v_x t = \ell \quad \longrightarrow \quad t=\frac{\ell}{v_x}
\\ \quad \quad \ell = 1.5m\]

The time is takes for the rock to travel the distance D into the catcher's hand.
\[ v_y t = D\]
Now we know the total Northernly distance "D" the rock has to travel by adding the original separating distance d (d=2m) to the distance person A travels during the flight of the rock.
\[D=d+v_at\]
so
\[v_yt = d+v_at\]
\[(v_y-v_a)t=d\]
\[t=\frac{d}{(v_y-v_a)}\]
Looking at that expression, we also know that the "y" velocity (it's Northern velocity) will be equivalent to person B's velocity, since the rock is only given extra velocity horizontally West.
\[t=\frac{d}{(v_b-v_a)}\]

We can now equate the two expressions for time

\[\frac{\ell}{v_x}=\frac{d}{(v_b-v_a)}\]
And solve for Vx - the horizontal (West) component of the rock's velocity