College Physics Quiz: Reference Frames And Relative Motion

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Reference Frames And Relative MotionQuestion 1 of 20

Two cars approach each other on a straight road. Car A travels north at $25 \text{ m/s}$ and car B travels south at $15 \text{ m/s}$ , both relative to the ground. What is the velocity of car A as observed from car B?

10 \text{ m/s}

north

40 \text{ m/s}

north

10 \text{ m/s}

south

40 \text{ m/s}

south

25 \text{ m/s}

north

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College Physics Quiz

College Physics Quiz: Reference Frames And Relative Motion

Practice Reference Frames And Relative Motion in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Reference Frames And Relative Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$10 \text{ m/s}$ north
$40 \text{ m/s}$ north (correct answer)
$10 \text{ m/s}$ south
$40 \text{ m/s}$ south
$25 \text{ m/s}$ north

Explanation: When you encounter relative velocity problems, you're dealing with how motion appears from different reference frames. The key insight is that velocities add vectorially when observers are moving relative to each other. To find car A's velocity as observed from car B, you need to subtract car B's velocity from car A's velocity. First, establish a coordinate system: let's say north is positive and south is negative. Car A moves at

+25 \text{ m/s}

and car B moves at

-15 \text{ m/s}

. The relative velocity formula is:

\vec{v}_{A \text{ relative to } B} = \vec{v}_A - \vec{v}_B = 25 - (-15) = 25 + 15 = 40 \text{ m/s}

north. From car B's perspective, car A approaches at 40 m/s northward. Looking at the wrong answers: Choice A (10 m/s north) incorrectly subtracts the speeds directly:

25 - 15 = 10

, ignoring that the cars move in opposite directions. Choice C (10 m/s south) makes the same subtraction error but also gets the direction wrong. Choice D (40 m/s south) correctly calculates the magnitude but reverses the direction—this would be car B's velocity relative to car A. Remember that when objects move toward each other, their relative speed is the sum of their individual speeds, not the difference. Always define your coordinate system clearly and be careful with signs when dealing with opposite directions. This concept appears frequently in collision and pursuit problems.

Question 2

A swimmer aims to cross a river by swimming directly toward the opposite shore. The swimmer can swim at $3 \text{ m/s}$ in still water, and the river current flows at $4 \text{ m/s}$ . What happens to the swimmer?

The swimmer reaches the opposite shore directly across from the starting point
The swimmer is carried downstream but still reaches the opposite shore
The swimmer is carried downstream and cannot reach the opposite shore (correct answer)
The swimmer moves upstream relative to the starting point
The swimmer remains stationary relative to the shore

Explanation: When you encounter river-crossing problems, you're dealing with vector addition of velocities. The key insight is that the swimmer's motion results from combining two independent velocity vectors: their swimming velocity and the river current velocity. Let's analyze what happens when the swimmer aims directly across the river. The swimmer moves at 3 m/s perpendicular to the shore, while the current pushes them downstream at 4 m/s parallel to the shore. These velocities add vectorially, creating a resultant velocity. The magnitude of this resultant is

\sqrt{3^2 + 4^2} = 5

m/s, directed downstream at an angle to the shore. Here's the critical issue: since the current (4 m/s) is faster than the swimmer's speed (3 m/s), the downstream component dominates the motion. The swimmer will be swept downstream faster than they can make progress across the river. They'll never reach the opposite shore because they're fighting a losing battle against the stronger current. Answer A is wrong because the swimmer doesn't travel in a straight line across—the current deflects their path. Answer B incorrectly assumes the swimmer can overcome the current's effect and still reach the other side. Answer D is impossible since the current flows downstream, not upstream. Study tip: In river problems, always compare the perpendicular swimming speed to the parallel current speed. If the current exceeds the swimmer's cross-river component, crossing becomes impossible. Draw vector diagrams to visualize these problems—they make the physics much clearer.

Question 3

Rain falls vertically downward at $5 \text{ m/s}$ relative to the ground. A person runs horizontally at $3 \text{ m/s}$ through the rain. At what angle from the vertical does the rain appear to fall from the person's perspective?

$31°$ (correct answer)
$37°$
$53°$
$59°$
$90°$

Explanation: When you encounter problems involving relative motion, think about how motion appears different from different reference frames. The key insight is that velocities add vectorially - you need to consider both the actual motion and the observer's motion. From the running person's perspective, the rain appears to have both a vertical component (its actual downward motion) and a horizontal component (because the person is moving horizontally through it). The rain's velocity relative to the person has a vertical component of

5 \text{ m/s}

downward and a horizontal component of

3 \text{ m/s}

in the direction opposite to the person's motion. To find the angle from vertical, use trigonometry. The horizontal component is

3 \text{ m/s}

and the vertical component is

5 \text{ m/s}

. The angle

\theta

from vertical satisfies:

\tan \theta = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{3}{5} = 0.6

Taking the inverse tangent:

\theta = \arctan(0.6) ≈ 31°

Choice A (

31°

) is correct. Choice B (

37°

) would result from confusing this with a 3-4-5 triangle and using the wrong angle. Choice C (

53°

) comes from calculating the angle from horizontal instead of vertical (

90° - 37° = 53°

). Choice D (

59°

) might result from computational errors or incorrect trigonometric relationships. Remember: in relative motion problems, always identify what each observer sees, then use vector addition. When finding angles, be careful about whether the problem asks for the angle from vertical or horizontal.

Question 4

A projectile is launched horizontally from a moving platform. The platform moves east at $5 \text{ m/s}$ and the projectile is launched north at $12 \text{ m/s}$ relative to the platform. What is the projectile's initial speed relative to the ground?

$7 \text{ m/s}$
$12 \text{ m/s}$
$13 \text{ m/s}$ (correct answer)
$17 \text{ m/s}$
$144 \text{ m/s}$

Explanation: When you encounter projectile motion problems involving moving platforms, you're dealing with relative velocity - the key is understanding that velocities add vectorially, not algebraically. The projectile has two velocity components relative to the ground: the platform's eastward motion (

5 \text{ m/s}

) and the projectile's northward launch velocity relative to the platform (

12 \text{ m/s}

). Since these directions are perpendicular, you need to use the Pythagorean theorem to find the resultant velocity. The projectile's speed relative to the ground is:

v = \sqrt{v_{east}^2 + v_{north}^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ m/s}

Looking at the wrong answers: Choice A (

7 \text{ m/s}

) represents the difference between the two velocities (

12 - 5

), which would only apply if the motions were in opposite directions along the same line. Choice B (

12 \text{ m/s}

) ignores the platform's motion entirely, giving only the launch velocity relative to the platform. Choice D (

17 \text{ m/s}

) incorrectly adds the velocities algebraically (

5 + 12

), treating them as if they're in the same direction rather than perpendicular. Remember: whenever you have perpendicular velocity components, always use vector addition with the Pythagorean theorem. If the problem involves velocities at other angles, you'll need to break them into components first, then combine using vector methods. Watch for the distinction between "relative to the platform" and "relative to the ground."

Question 5

In a reference frame accelerating upward at $2 \text{ m/s}^2$ , what is the apparent acceleration of a freely falling object that would fall at $9.8 \text{ m/s}^2$ downward in an inertial frame?

$7.8 \text{ m/s}^2$ downward
$9.8 \text{ m/s}^2$ downward
$11.8 \text{ m/s}^2$ downward (correct answer)
$2 \text{ m/s}^2$ upward
$0 \text{ m/s}^2$

Explanation: When analyzing motion in non-inertial (accelerating) reference frames, you need to account for fictitious forces that appear due to the frame's acceleration. This is a key concept in mechanics that often confuses students. To find the apparent acceleration, you must consider how the object's motion appears relative to the accelerating reference frame. The freely falling object has a real acceleration of

9.8 \text{ m/s}^2

downward due to gravity. However, since you're observing from a frame accelerating upward at

2 \text{ m/s}^2

, the object appears to accelerate even faster downward relative to your frame. The apparent acceleration equals the object's real acceleration minus the reference frame's acceleration (accounting for directions). Since upward is positive: apparent acceleration =

-9.8 - (+2) = -11.8 \text{ m/s}^2

, or

11.8 \text{ m/s}^2

downward. This makes C correct. Option A (

7.8 \text{ m/s}^2

downward) incorrectly subtracts the frame's acceleration instead of adding it, representing the common error of wrong sign convention. Option B (

9.8 \text{ m/s}^2

downward) ignores the reference frame's acceleration entirely, treating it as if you're in an inertial frame. Option D (

2 \text{ m/s}^2

upward) only considers the frame's acceleration while ignoring gravity completely. Remember this pattern: when your reference frame accelerates in one direction, objects appear to accelerate more strongly in the opposite direction. Always add the magnitudes when the accelerations are in opposite directions.

Question 6

A ball rolls across the floor of a train car. The train accelerates forward at $1.5 \text{ m/s}^2$ , and the ball accelerates backward at $2 \text{ m/s}^2$ relative to the train. What is the ball's acceleration relative to the ground?

$0.5 \text{ m/s}^2$ backward (correct answer)
$0.5 \text{ m/s}^2$ forward
$2 \text{ m/s}^2$ backward
$3.5 \text{ m/s}^2$ forward
$3.5 \text{ m/s}^2$ backward

Explanation: When you encounter problems involving motion in multiple reference frames, you need to carefully apply vector addition to relate accelerations between different observers. This is a classic relative motion problem where the ball's motion must be analyzed from both the train's perspective and the ground's perspective. Let's establish our coordinate system with forward as positive. The train accelerates forward at

+1.5 \text{ m/s}^2

relative to the ground. The ball accelerates backward at

-2 \text{ m/s}^2

relative to the train. To find the ball's acceleration relative to the ground, you add these accelerations vectorially:

a_{ball,ground} = a_{ball,train} + a_{train,ground} = (-2) + (+1.5) = -0.5 \text{ m/s}^2

The negative sign indicates backward motion, so the ball accelerates at

0.5 \text{ m/s}^2

backward relative to the ground. Looking at the wrong answers: B gives the correct magnitude but wrong direction—this comes from incorrectly treating backward as positive. C ignores the train's motion entirely, using only the ball's acceleration relative to the train. D incorrectly adds the magnitudes without considering directions, giving

2 + 1.5 = 3.5

and assuming forward motion. The key strategy for relative motion problems is to always define your positive direction clearly and remember that accelerations add as vectors. When an object moves in one reference frame that's itself accelerating, you must account for both motions to find the acceleration relative to a stationary observer.

Question 7

A river flows east at $3 \text{ m/s}$ . A boat travels from the north bank to the south bank, taking the shortest possible time. If the boat can travel at $4 \text{ m/s}$ relative to the water, what is the boat's velocity relative to the riverbank during this crossing?

$3 \text{ m/s}$ east
$4 \text{ m/s}$ south
$5 \text{ m/s}$ southeast (correct answer)
$7 \text{ m/s}$ southeast
$1 \text{ m/s}$ south

Explanation: This is a relative velocity problem where you need to combine the boat's motion through water with the water's motion relative to the shore. When the problem asks for the "shortest possible time" to cross, the boat should aim directly south relative to the water, ignoring the eastward drift caused by the current. To find the boat's velocity relative to the riverbank, you add the velocity vectors. The boat moves at 4 m/s south relative to the water, while the water flows at 3 m/s east relative to the shore. These velocities are perpendicular, so you can use the Pythagorean theorem:

v = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \text{ m/s}

. The direction is southeast since the boat has both southward and eastward components. Answer A (3 m/s east) only accounts for the river's current, ignoring the boat's southward motion entirely. Answer B (4 m/s south) represents the boat's velocity relative to the water, not relative to the shore—it ignores how the current affects the boat's actual path. Answer D (7 m/s southeast) incorrectly adds the velocity magnitudes directly (3 + 4 = 7) rather than using vector addition for perpendicular components. Remember that "shortest time" means pointing directly toward your destination relative to the medium you're moving through (the water), not relative to the ground. Always use vector addition when combining velocities, and watch for the Pythagorean theorem when dealing with perpendicular velocity components.

Question 8

A train travels east at $30 \text{ m/s}$ relative to the ground. A passenger walks west through the train at $2 \text{ m/s}$ relative to the train. What is the velocity of the passenger relative to the ground?

$28 \text{ m/s}$ east (correct answer)
$32 \text{ m/s}$ east
$28 \text{ m/s}$ west
$32 \text{ m/s}$ west
$2 \text{ m/s}$ west

Explanation: When you encounter relative velocity problems, you're dealing with how motion appears from different reference frames. The key is to carefully track which velocities are measured relative to what, then use vector addition to find the final result. Let's establish a coordinate system where east is positive and west is negative. The train moves at

+30 \text{ m/s}

relative to the ground (eastward). The passenger walks at

-2 \text{ m/s}

relative to the train (westward). To find the passenger's velocity relative to the ground, you add these vectors:

v_{passenger/ground} = v_{passenger/train} + v_{train/ground} = (-2) + (+30) = +28 \text{ m/s}

. This means

28 \text{ m/s}

eastward. Looking at the wrong answers: Choice B (

32 \text{ m/s}

east) represents the common error of adding the magnitudes without considering direction—you'd get this by incorrectly calculating

30 + 2 = 32

. Choice C (

28 \text{ m/s}

west) has the correct magnitude but wrong direction, suggesting confusion about the sign conventions or vector addition. Choice D (

32 \text{ m/s}

west) combines both errors: wrong magnitude and wrong direction. The correct answer is A:

28 \text{ m/s}

east. Study tip: Always establish a clear coordinate system first (positive/negative directions), convert all velocities to that system, then add algebraically. Remember that "relative to" problems require vector addition:

\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}

. Practice identifying which reference frame each velocity is measured from.

Question 9

An airplane flies due north at $200 \text{ km/h}$ relative to the air. A wind blows from west to east at $50 \text{ km/h}$ . What is the direction of the airplane's velocity relative to the ground?

Due north
$14°$ east of north (correct answer)
$14°$ west of north
$76°$ east of north
Due east

Explanation: This is a classic vector addition problem involving relative velocity. When an object moves through a medium that's also moving (like a plane flying through moving air), you must add the vectors to find the motion relative to the ground. The airplane flies north at 200 km/h relative to the air, while wind blows east at 50 km/h. To find the plane's ground velocity, treat these as perpendicular vectors forming a right triangle. The plane's airspeed forms the northward leg (200 km/h), the wind forms the eastward leg (50 km/h), and the resultant ground velocity is the hypotenuse. The direction angle east of north is found using:

\tan(\theta) = \frac{\text{eastward component}}{\text{northward component}} = \frac{50}{200} = 0.25

Therefore:

\theta = \arctan(0.25) = 14°

The airplane's velocity is 14° east of north, making (B) correct. (A) assumes the wind has no effect, ignoring that the eastward wind component pushes the plane off its due-north heading. (C) gets the angle magnitude right but reverses the direction—an eastward wind pushes the plane east, not west, of its intended path. (D) likely comes from incorrectly using

\arctan(200/50)

or confusing which vector component goes in the numerator versus denominator. Study tip: In relative velocity problems, always draw a vector diagram with the object's velocity and the medium's velocity as perpendicular sides of a right triangle. The angle is measured from the larger velocity vector toward the smaller one.

Question 10

A person throws a ball horizontally from the deck of a ship moving at constant velocity. Neglecting air resistance, what path does the ball follow as observed from the ship?

A straight horizontal line at constant height
A straight vertical line downward (correct answer)
A parabolic path curving forward and downward
A parabolic path curving backward and downward
A circular arc

Explanation: This question tests your understanding of relative motion and reference frames in physics. When analyzing motion, the path an object follows depends entirely on your point of reference. From the ship's reference frame, you and the ball initially share the same horizontal velocity as the ship. When you throw the ball horizontally, you're not adding any additional horizontal velocity relative to the ship - you're simply releasing it. Since the ship maintains constant velocity and there's no air resistance, the ball continues moving horizontally at the same speed as the ship throughout its flight. However, gravity still acts on the ball, pulling it downward with acceleration

g = 9.8 \text{ m/s}^2

. Since there's no horizontal acceleration relative to the ship, the ball only accelerates vertically downward. From your perspective on the ship, the ball appears to fall straight down. Looking at the wrong answers: A) suggests the ball maintains constant height, ignoring gravity's effect entirely. C) describes a forward-curving parabola, which might occur if you threw the ball with additional forward velocity relative to the ship, but that's not the case here. D) suggests backward curvature, which would only happen if some force (like air resistance) opposed the ship's motion - but we're told to neglect air resistance. Strategy tip: In relative motion problems, always establish your reference frame first. Motion that appears parabolic from one reference frame (like the ground) can appear completely different from another reference frame (like the moving ship). The key is identifying which forces act in your chosen reference frame.

Question 11

Two trains travel on parallel tracks. Train A moves east at $60 \text{ km/h}$ and train B moves west at $40 \text{ km/h}$ , both relative to the ground. A passenger on train A observes train B. What is the rate at which the distance between the trains is changing?

$20 \text{ km/h}$
$50 \text{ km/h}$
$100 \text{ km/h}$ (correct answer)
$120 \text{ km/h}$
$2400 \text{ km/h}$

Explanation: When you encounter problems involving objects moving in opposite directions, you're dealing with relative velocity and the rate of separation. The key insight is that when two objects move away from each other, their relative speeds add together. Since train A moves east at

60 \text{ km/h}

and train B moves west at

40 \text{ km/h}

, they're traveling in opposite directions. To find how quickly the distance between them changes, you add their speeds:

60 + 40 = 100 \text{ km/h}

. This represents the rate at which they separate from each other. Think of it this way: in one hour, train A will be 60 km farther east from its starting point, while train B will be 40 km farther west from its starting point. The total increase in distance between them is

60 + 40 = 100

km. Option A (

20 \text{ km/h}

) incorrectly subtracts the speeds, which you'd only do if the trains were moving in the same direction with one catching up to the other. Option B (

50 \text{ km/h}

) represents the average of the two speeds, which has no physical meaning in this context. Option D (

120 \text{ km/h}

) incorrectly adds both speeds twice or misapplies the relative velocity concept. Remember this pattern: when objects move in opposite directions, add their speeds to find the rate of separation. When they move in the same direction, subtract the slower speed from the faster speed to find the rate one gains on the other.

Question 12

A person stands on a moving walkway that travels east at $1.5 \text{ m/s}$ . The person walks west on the walkway at $0.8 \text{ m/s}$ relative to the walkway. After $10 \text{ s}$ , how far has the person moved relative to the ground?

$7 \text{ m}$ east (correct answer)
$7 \text{ m}$ west
$8 \text{ m}$ east
$15 \text{ m}$ east
$23 \text{ m}$ east

Explanation: When you encounter relative motion problems, you need to carefully track velocities in the same reference frame. Here, you're asked for the person's motion relative to the ground, so you must combine the walkway's motion and the person's motion relative to the walkway. The walkway moves east at

1.5 \text{ m/s}

relative to the ground. The person walks west at

0.8 \text{ m/s}

relative to the walkway. To find the person's velocity relative to the ground, you add these velocities vectorially. Taking east as positive: walkway velocity =

+1.5 \text{ m/s}

, person's velocity relative to walkway =

-0.8 \text{ m/s}

(west is negative). The person's velocity relative to ground =

1.5 + (-0.8) = 0.7 \text{ m/s}

east. After

10 \text{ s}

, the displacement is:

d = vt = 0.7 \times 10 = 7 \text{ m}

east. Choice A (

7 \text{ m}

east) correctly combines both motions. Choice B (

7 \text{ m}

west) gets the magnitude right but assumes the person's walking speed dominates the direction. Choice C (

8 \text{ m}

east) likely adds the speeds incorrectly (

1.5 - 0.8 = 0.8

, then

0.8 \times 10 = 8

). Choice D (

15 \text{ m}

east) incorrectly adds the magnitudes (

1.5 + 0.8 = 2.3

, roughly

15

after

10

seconds) without considering opposite directions. Always establish a coordinate system first, then add velocities algebraically, paying careful attention to signs. The final motion is always relative to your chosen reference frame—here, the ground.

Question 13

An escalator moves upward at $2 \text{ m/s}$ . A person walks down the escalator at $3 \text{ m/s}$ relative to the escalator steps. What is the person's velocity relative to the building?

$1 \text{ m/s}$ upward
$1 \text{ m/s}$ downward (correct answer)
$2 \text{ m/s}$ upward
$3 \text{ m/s}$ downward
$5 \text{ m/s}$ downward

Explanation: When you encounter problems involving relative motion, you need to carefully consider which reference frame you're measuring velocities against. This problem requires you to find the person's velocity relative to the building (ground), given their motion relative to the escalator. To solve this, use vector addition of velocities. The escalator moves upward at

2 \text{ m/s}

relative to the building. The person walks downward at

3 \text{ m/s}

relative to the escalator steps. To find the person's velocity relative to the building, add these velocities vectorially. Setting upward as positive: the escalator's velocity is

+2 \text{ m/s}

, and the person's velocity relative to the escalator is

-3 \text{ m/s}

(downward). The person's velocity relative to the building equals:

2 + (-3) = -1 \text{ m/s}

, which means

1 \text{ m/s}

downward. Choice A (

1 \text{ m/s}

upward) incorrectly adds the magnitudes without considering direction, giving

1 \text{ m/s}

but in the wrong direction. Choice C (

2 \text{ m/s}

upward) mistakenly uses only the escalator's velocity, ignoring the person's walking motion. Choice D (

3 \text{ m/s}

downward) incorrectly uses only the person's velocity relative to the escalator, ignoring the escalator's upward motion entirely. The correct answer is B:

1 \text{ m/s}

downward. Study tip: Always identify your reference frames clearly in relative motion problems. Write down each velocity with its proper sign (direction), then add algebraically. The key equation is:

\vec{v}_{person,ground} = \vec{v}_{person,escalator} + \vec{v}_{escalator,ground}

Question 14

A car travels around a circular track at constant speed. From the reference frame of the car, which force must be present to explain the car's motion?

A centripetal force directed toward the center of the track
A centrifugal force directed away from the center of the track (correct answer)
A tangential force in the direction of motion
No additional forces beyond those in an inertial frame
A gravitational force stronger than normal

Explanation: When analyzing circular motion problems, the key insight is recognizing that different reference frames require different explanations for the same physical situation. From an inertial (stationary) reference frame, a car moving in a circle experiences centripetal acceleration toward the center, requiring a real centripetal force (friction between tires and road) to maintain circular motion. However, the car's reference frame is non-inertial because it's accelerating, which fundamentally changes how we must explain the motion. In the car's reference frame, the passengers feel pushed outward against their seats, even though the car maintains constant circular motion. To explain this sensation and the apparent "equilibrium" in this accelerating frame, we must introduce a fictitious force called centrifugal force, directed away from the center. This pseudo-force balances the inward centripetal force, explaining why objects in the car don't appear to accelerate toward the center from the car's perspective. Choice A is incorrect because centripetal force explains motion from an inertial frame, not the car's frame. Choice C is wrong because tangential forces would change the car's speed, but we're told speed is constant. Choice D fails because non-inertial frames always require additional fictitious forces to explain observed motion. Remember: fictitious forces like centrifugal force only appear in accelerating reference frames. When you're asked about motion "from the reference frame" of an accelerating object, look for these pseudo-forces rather than the real forces that external observers would identify.

Question 15

A ball is dropped from a train moving horizontally at constant velocity $v$ . Which statement best describes the ball's motion as observed from the ground?

The ball falls straight down with no horizontal motion
The ball follows a parabolic path with initial horizontal velocity $v$ (correct answer)
The ball follows a parabolic path with constant horizontal acceleration
The ball moves in a straight line at angle to the vertical
The ball's horizontal velocity decreases as it falls

Explanation: This question tests your understanding of projectile motion and reference frames. When analyzing motion, you must carefully consider what initial conditions the object has when it begins its trajectory. When the ball is dropped from the moving train, it inherits the train's horizontal velocity

v

at the moment of release. This is a crucial principle: objects maintain the motion they had before being released (Newton's first law). Since there's no horizontal force acting on the ball after release (ignoring air resistance), it continues moving horizontally at velocity

v

while simultaneously accelerating downward due to gravity at

g = 9.8 \text{ m/s}^2

. The combination of constant horizontal velocity and increasing downward velocity creates the characteristic parabolic trajectory of projectile motion. Answer B correctly captures both elements: the parabolic path and the initial horizontal velocity

v

. Answer A is wrong because it ignores the horizontal motion inherited from the train. Answer C incorrectly suggests horizontal acceleration—but once released, the ball experiences no horizontal forces, so its horizontal velocity remains constant. Answer D describes motion along a straight diagonal line, which would only occur if the ball had both horizontal and vertical velocity components that remained in constant proportion, not the case here where only the vertical component changes. Remember this key insight for projectile problems: objects always retain the velocity they had at the moment of release. Whether it's a ball dropped from a moving train or a package dropped from an airplane, the initial horizontal motion continues unchanged while gravity acts independently in the vertical direction.

Question 16

A boat travels at $8 \text{ m/s}$ relative to still water. The boat heads directly across a river that flows east at $6 \text{ m/s}$ . What is the magnitude of the boat's velocity relative to the riverbank?

$2 \text{ m/s}$
$7 \text{ m/s}$
$10 \text{ m/s}$ (correct answer)
$14 \text{ m/s}$
$48 \text{ m/s}$

Explanation: When you encounter problems involving relative motion, especially with boats and rivers, you're dealing with vector addition. The key insight is that velocities add as vectors, not simple numbers, because they have both magnitude and direction. Here, the boat moves at 8 m/s relative to the water (let's say northward across the river), while the river current carries everything eastward at 6 m/s. The boat's actual motion relative to the riverbank is the vector sum of these two velocities. Since the boat heads directly across (perpendicular to the current), these velocity vectors form a right triangle. Using the Pythagorean theorem:

v_{resultant} = \sqrt{v_{boat}^2 + v_{river}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ m/s}

Choice A (2 m/s) represents the common error of subtracting the velocities, treating them as opposing forces rather than perpendicular vectors. Choice B (7 m/s) might result from incorrectly averaging the two speeds or from computational errors. Choice D (14 m/s) comes from simply adding the magnitudes, ignoring that these are perpendicular vectors, not collinear ones. Remember: when dealing with perpendicular relative motion problems, always visualize the vector triangle and apply the Pythagorean theorem. If the velocities were parallel (same or opposite directions), you'd add or subtract directly, but perpendicular motion requires vector addition using the Pythagorean theorem.

Question 17

Two boats start from the same point. Boat A travels northeast at $10 \text{ m/s}$ and boat B travels northwest at $10 \text{ m/s}$ , both relative to the water. If there is no current, what is the velocity of boat A relative to boat B?

$10 \text{ m/s}$ east
$10\sqrt{2} \text{ m/s}$ east
$14.1 \text{ m/s}$ east (correct answer)
$20 \text{ m/s}$ northeast
$0 \text{ m/s}$

Explanation: This is a relative velocity problem that requires vector analysis. When finding the velocity of one object relative to another, you need to subtract their velocity vectors, not just their magnitudes. To find boat A's velocity relative to boat B, calculate

\vec{v}_{A} - \vec{v}_{B}

. Set up a coordinate system where east is positive x and north is positive y. Boat A travels northeast at

10 \text{ m/s}

, so its components are

\vec{v}_{A} = (10\cos45°, 10\sin45°) = (7.07, 7.07) \text{ m/s}

. Boat B travels northwest at

10 \text{ m/s}

, so

\vec{v}_{B} = (-10\cos45°, 10\sin45°) = (-7.07, 7.07) \text{ m/s}

. The relative velocity is

\vec{v}_{A} - \vec{v}_{B} = (7.07, 7.07) - (-7.07, 7.07) = (14.14, 0) \text{ m/s}

. This gives a magnitude of

14.14 \text{ m/s}

due east, which equals

14.1 \text{ m/s}

when rounded. Choice A incorrectly uses just the magnitude of one boat's velocity. Choice B gives

10\sqrt{2} \approx 14.1 \text{ m/s}

but fails to recognize this comes from proper vector subtraction, not some arbitrary calculation. Choice D incorrectly adds the magnitudes and assumes the direction splits the difference between northeast and northwest. Study tip: In relative velocity problems, always work with vector components rather than trying to use magnitudes directly. The relative velocity

\vec{v}_{A \text{ rel } B} = \vec{v}_{A} - \vec{v}_{B}

requires vector subtraction, which means subtracting corresponding components.

Question 18

A boat travels upstream in a river at $8.0 \text{ m/s}$ relative to the riverbank. The same boat travels downstream at $18 \text{ m/s}$ relative to the riverbank. A log floating in the river passes a fixed bridge. How long after the log passes the bridge will the downstream-traveling boat pass the same bridge if the boat and log are initially $520 \text{ m}$ apart?

$26 \text{ s}$
$29 \text{ s}$
$40 \text{ s}$ (correct answer)
$52 \text{ s}$

Explanation: Let

v_b

= boat speed in still water,

v_r

= river speed. Then

v_b - v_r = 8.0

and

v_b + v_r = 18

. Solving:

v_b = 13 \text{ m/s}

v_r = 5.0 \text{ m/s}

. The log moves at river speed

5.0 \text{ m/s}

. The boat (downstream) moves at

18 \text{ m/s}

. Their relative speed is

18 - 5 = 13 \text{ m/s}

. Time to close

520 \text{ m}

gap:

t = 520/13 = 40 \text{ s}

. Other choices use incorrect relative speeds or wrong interpretations of the reference frame problem.

Question 19

A train traveling at $30 \text{ m/s}$ passes a stationary observer on a platform. At the moment the front of the train passes the observer, a passenger at the back of the $150 \text{ m}$ long train throws a ball forward at $20 \text{ m/s}$ relative to the train. How long after the front of the train passes the platform observer does the ball pass the same observer?

$2.5 \text{ s}$
$3.0 \text{ s}$ (correct answer)
$5.0 \text{ s}$
$7.5 \text{ s}$

Explanation: The ball's velocity relative to the platform observer is

30 + 20 = 50 \text{ m/s}

. When thrown, the ball is

150 \text{ m}

behind the front of the train. The ball must travel this

150 \text{ m}

distance at

50 \text{ m/s}

to reach the observer's position, taking

t = 150/50 = 3.0 \text{ s}

. Choice A uses only the train's speed. Choice C incorrectly uses the ball's speed relative to the train. Choice D adds the times incorrectly.

Question 20

Two trains travel on parallel tracks in opposite directions. Train A moves at $25 \text{ m/s}$ east, and train B moves at $35 \text{ m/s}$ west. A passenger on train A observes train B for $3.0 \text{ s}$ as it passes. What is the length of train B?

$75 \text{ m}$
$105 \text{ m}$
$180 \text{ m}$ (correct answer)
$240 \text{ m}$

Explanation: The relative speed between the trains is

25 + 35 = 60 \text{ m/s}

since they move in opposite directions. Train B passes completely by train A's observer in

3.0 \text{ s}

, so train B's length is

60 \times 3.0 = 180 \text{ m}

. Choice A uses only train A's speed. Choice B uses only train B's speed. Choice D incorrectly considers both train lengths or uses wrong relative motion concepts.