Let east be the positive x-direction ($$\hat{i}$$) and north be the positive y-direction ($$\hat{j}$$). Then $$v_A = 30\hat{i}$$ m/s and $$v_B = 40\hat{j}$$ m/s. The velocity of A relative to B is $$v_{AB} = v_A - v_B = 30\hat{i} - 40\hat{j}$$ m/s. The magnitude (speed) is $$|v_{AB}| = \sqrt{30^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$$ m/s. The direction is given by $$\theta = \arctan(\frac{-40}{30}) \approx -53°$$, which corresponds to 53° South of East.

Let the velocity of the rain relative to the ground be $$v_{RG} = -8\hat{j}$$ m/s and the velocity of the cyclist relative to the ground be $$v_{CG} = 6\hat{i}$$ m/s. The velocity of the rain relative to the cyclist is $$v_{RC} = v_{RG} - v_{CG} = -6\hat{i} - 8\hat{j}$$ m/s. The angle $$ heta$$ from the vertical is given by $$\tan(\theta) = \frac{|v_x|}{|v_y|} = \frac{6}{8} = \frac{3}{4}$$. Therefore, the angle is $$\theta = \arctan(3/4)$$ from the vertical.

Let $$v_{PG}$$ be the velocity of the passenger relative to the ground, $$v_{PT}$$ be the velocity of the passenger relative to the train, and $$v_{TG}$$ be the velocity of the train relative to the ground. Using the relative velocity addition formula, $$v_{PG} = v_{PT} + v_{TG}$$. Since both velocities are in the same direction (east), we can add their magnitudes: $$v_{PG} = 2 \text{ m/s} + 20 \text{ m/s} = 22 \text{ m/s}$$ due east.

The definition of an inertial reference frame is a frame in which Newton's first law (the law of inertia) holds true. This means that an object experiences no net force will not accelerate; it will maintain a constant velocity. Distractor A is incorrect because both objects at rest and at constant velocity have zero net force. Distractor B is incorrect because objects can accelerate in an inertial frame if a net force acts on them. Distractor D is partially correct but incorrectly states that other constant velocity frames are not inertial.

The velocity of the boat relative to the ground ($$v_{BG}$$) must be purely east. This velocity is the vector sum of the boat's velocity relative to the water ($$v_{BW}$$) and the water's velocity relative to the ground ($$v_{WG}$$). To counteract the southward current ($$v_{WG} = -2.0\hat{j}$$ m/s), the boat must have a northward component of velocity relative to the water. Let $$ heta$$ be the angle North of East. Then $$v_{BW} = (4.0 \cos\theta)\hat{i} + (4.0 \sin\theta)\hat{j}$$. The y-component of $$v_{BG}$$ must be zero: $$(4.0 \sin\theta) - 2.0 = 0$$. This gives $$\sin\theta = 2.0/4.0 = 0.5$$, so $$ heta = 30°$$. The direction is 30° North of East.

When the coin is dropped, it has the same initial horizontal velocity as the cruise ship. Assuming air resistance is negligible, there is no horizontal force acting on the coin. According to Newton's first law, the coin's horizontal velocity will remain constant. Therefore, the observer on the dock sees the coin move horizontally at the ship's constant velocity while it accelerates vertically downwards.

This question tests AP Physics C Mechanics skills: understanding reference frames and relative motion. A reference frame is a perspective from which motion is measured. Relative motion denotes how an object moves in relation to a chosen frame. In this scenario, Car A moves east at 30 m/s and Car B moves east at 25 m/s, both in the same direction, and we need their relative speed. Choice B is correct because the relative speed between objects moving in the same direction is the difference of their speeds: 30 - 25 = 5 m/s. Choice A is incorrect because it adds the speeds (30 + 25 = 55 m/s), which would only be correct if the cars moved in opposite directions. To help students: Emphasize that relative speed for same-direction motion involves subtraction, not addition. Use the concept of 'catching up' - Car A gains on Car B at 5 m/s.

This question tests AP Physics C Mechanics skills: understanding reference frames and relative motion. A reference frame is a perspective from which motion is measured. Relative motion denotes how an object moves in relation to a chosen frame. In this scenario, the train moves east at 20 m/s while the passenger walks east at 2 m/s inside, and we need the passenger's velocity relative to the train. Choice C is correct because relative to the train frame, the passenger moves at 2 m/s east - this is exactly the walking speed given in the problem. Choice B is incorrect because it adds the velocities (20 + 2 = 22 m/s), which would give the passenger's velocity relative to the ground, not the train. To help students: Emphasize that 'relative to the train' means using the train as the reference frame where the train itself is stationary. Draw diagrams showing the same motion from different reference frames to build intuition.

This question tests AP Physics C Mechanics skills: understanding reference frames and relative motion. A reference frame is a perspective from which motion is measured. Relative motion denotes how an object moves in relation to a chosen frame. In this scenario, Plane A flies east at 250 m/s and Plane B flies east at 230 m/s, and we need Plane A's speed relative to Plane B. Choice B is correct because from Plane B's reference frame, Plane A appears to move at 250 - 230 = 20 m/s east, which is how fast A gains on B. Choice A is incorrect because it adds the speeds (250 + 230 = 480 m/s), which would only apply if the planes moved in opposite directions. To help students: Use the concept of one plane as a 'moving platform' - from B's cockpit, how fast does A appear to approach? Practice problems with objects moving in the same direction at different speeds.

This question tests AP Physics C Mechanics skills: understanding reference frames and relative motion. A reference frame is a perspective from which motion is measured. Relative motion denotes how an object moves in relation to a chosen frame. In this scenario, the train moves east at 15 m/s while the passenger walks west at 3 m/s inside the train, and we need the passenger's velocity relative to the ground. Choice A is correct because from the ground frame, we must add the train's velocity (15 m/s east) to the passenger's velocity relative to the train (3 m/s west = -3 m/s east), giving 15 + (-3) = 12 m/s east. Choice B is incorrect because it subtracts in the wrong direction (15 + 3 = 18), assuming the passenger walks east instead of west. To help students: Use vector addition with clear sign conventions - establish east as positive. Practice problems where passengers move opposite to the vehicle's direction to reinforce proper vector addition.