College Physics Quiz: Motion In Two Or Three Dimensions

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Motion In Two Or Three DimensionsQuestion 1 of 20

A particle moves in the xy-plane with position given by $\vec{r}(t) = (4t^2, 3t^2)$ where distances are in meters and time in seconds. What is the magnitude of the particle's velocity at t = 2.0 s?

10 m/s

14 m/s

20 m/s

28 m/s

36 m/s

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College Physics Quiz

College Physics Quiz: Motion In Two Or Three Dimensions

Practice Motion In Two Or Three Dimensions 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 Motion In Two Or Three Dimensions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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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

A particle moves in the xy-plane with position given by $\vec{r}(t) = (4t^2, 3t^2)$ where distances are in meters and time in seconds. What is the magnitude of the particle's velocity at t = 2.0 s?

10 m/s
14 m/s
20 m/s (correct answer)
28 m/s
36 m/s

Explanation: When you encounter position vectors as functions of time, you're dealing with kinematics in multiple dimensions. The key insight is that velocity is the time derivative of position, so you need to differentiate each component separately. Given

\vec{r}(t) = (4t^2, 3t^2)

, find the velocity vector by taking the derivative:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (8t, 6t)

. At t = 2.0 s, this becomes

\vec{v}(2) = (16, 12)

m/s. The magnitude of velocity is found using the Pythagorean theorem:

|\vec{v}| = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

m/s. Let's examine why the other answers are incorrect. Choice A (10 m/s) would result if you incorrectly calculated

\sqrt{8^2 + 6^2}

using the velocity coefficients rather than evaluating at t = 2. Choice B (14 m/s) might come from adding the velocity components (16 + 12 = 28) then dividing by 2, which has no physical meaning. Choice D (28 m/s) results from simply adding the velocity components instead of finding the vector magnitude properly. Remember that for any vector quantity, the magnitude requires the square root of the sum of squared components. Don't confuse this with scalar addition. Also, always substitute your time value after differentiating—taking derivatives and evaluating at specific times are separate steps that must be done in the correct order.

Question 2

A particle moves in a circular path of radius 2.0 m with constant speed 8.0 m/s. What is the magnitude of its centripetal acceleration?

4.0 m/s²
16 m/s²
32 m/s² (correct answer)
64 m/s²
128 m/s²

Explanation: When you encounter circular motion problems, you're dealing with centripetal acceleration—the acceleration that keeps an object moving in a circular path. This acceleration always points toward the center of the circle and has a specific formula you need to master. For centripetal acceleration, use

a_c = \frac{v^2}{r}

, where

v

is the speed and

r

is the radius. With the given values of

v = 8.0 \text{ m/s}

and

r = 2.0 \text{ m}

a_c = \frac{(8.0)^2}{2.0} = \frac{64}{2.0} = 32 \text{ m/s}^2

This confirms answer C is correct. Now let's see where the wrong answers come from. Answer A (4.0 m/s²) likely results from incorrectly using

\frac{v}{r^2} = \frac{8.0}{(2.0)^2} = \frac{8.0}{4.0} = 2.0

, or perhaps

\frac{2v}{r} = \frac{16}{4} = 4.0

—both reflect formula confusion. Answer B (16 m/s²) comes from forgetting to square the velocity, calculating

\frac{v \cdot r}{r} = v = 8.0

2v = 16

. Answer D (64 m/s²) represents the numerator

v^2

without dividing by the radius—a common mistake when students remember part of the formula but not the complete calculation. Remember this key pattern: centripetal acceleration problems always use

a_c = \frac{v^2}{r}

. The velocity must be squared, and many wrong answers come from forgetting this step or misplacing the radius in the calculation.

Question 3

A projectile is launched from ground level at an angle θ above horizontal. Which statement about the projectile's motion is correct?

The horizontal component of velocity decreases linearly with time due to air resistance effects
The vertical component of velocity is zero at the launch point and maximum at the highest point
The projectile's speed is minimum at the highest point of the trajectory
The horizontal and vertical displacements are equal when the projectile returns to ground level
The time to reach maximum height equals the time to fall from maximum height to ground level (correct answer)

Explanation: When analyzing projectile motion, you need to separate the motion into horizontal and vertical components and consider how gravity affects each differently. At the highest point of a projectile's trajectory, the vertical component of velocity becomes zero (the projectile momentarily stops rising before falling). However, the horizontal component remains constant throughout the flight (assuming no air resistance). This means the projectile's total speed at the highest point equals just the horizontal component, making it the minimum speed during the entire flight. Let's examine why the other options are incorrect: A) The horizontal velocity component actually remains constant throughout the flight when air resistance is negligible (the standard assumption in introductory physics). Only gravity acts on the projectile, and gravity only affects vertical motion. B) This gets the vertical velocity behavior completely backwards. At launch, the vertical component starts at its maximum value (

v_0 \sin θ

) and decreases due to gravity until it reaches zero at the highest point. D) For a projectile launched and landing at the same height, the horizontal displacement (range) generally doesn't equal the vertical displacement, which is zero when it returns to ground level. The range depends on the launch angle, with maximum range occurring at 45°. Remember this key insight: at the highest point of any projectile's path, the vertical velocity is always zero, leaving only the horizontal component. This makes the highest point the location of minimum speed—a useful fact for quickly analyzing projectile motion problems.

Question 4

A ball is thrown horizontally from the top of a 45-meter tall building with an initial speed of 20 m/s. How far horizontally does the ball travel before hitting the ground? (Use g = 10 m/s²)

30 m
45 m
60 m (correct answer)
90 m
120 m

Explanation: Projectile motion problems involve analyzing horizontal and vertical motion separately. When an object is thrown horizontally, it has zero initial vertical velocity but maintains constant horizontal velocity throughout its flight. To find the horizontal distance, you need two pieces of information: the horizontal velocity and the time of flight. The horizontal velocity remains constant at 20 m/s. To find the time of flight, analyze the vertical motion using

h = \frac{1}{2}gt^2

(since initial vertical velocity is zero). Solving for time:

45 = \frac{1}{2}(10)t^2

, which gives

t^2 = 9

, so

t = 3

seconds. The horizontal distance is:

d = v_x \times t = 20 \times 3 = 60

meters. Looking at the wrong answers: Choice A (30 m) represents a common error where students might use only half the correct time or confuse the calculation. Choice B (45 m) occurs when students incorrectly use the building height as the horizontal distance, mixing up vertical and horizontal components. Choice D (90 m) suggests using an incorrect time value, possibly doubling the correct time or misapplying the kinematic equations. The key strategy for projectile motion is to always separate horizontal and vertical components. Use the vertical motion to find time (since gravity only affects vertical motion), then apply that time to the horizontal motion where velocity stays constant. Remember: horizontal motion is uniform, vertical motion is uniformly accelerated.

Question 5

A ball moves in a vertical circle of radius 0.8 m. At the bottom of the circle, its speed is 6.0 m/s. What is the magnitude of the ball's acceleration at this point? (Use g = 10 m/s²)

10 m/s²
35 m/s²
45 m/s²
55 m/s² (correct answer)
65 m/s²

Explanation: When analyzing circular motion problems, you need to consider both the centripetal acceleration (directed toward the center) and gravitational acceleration. At the bottom of a vertical circle, these accelerations add together since they point in the same direction. At the bottom of the circle, two accelerations act on the ball. The centripetal acceleration is

a_c = \frac{v^2}{r} = \frac{(6.0)^2}{0.8} = \frac{36}{0.8} = 45 \text{ m/s}^2

, directed upward toward the center. Gravitational acceleration is

g = 10 \text{ m/s}^2

, also directed upward (toward the center) at this position. The total acceleration magnitude is

a_{total} = a_c + g = 45 + 10 = 55 \text{ m/s}^2

. Choice A (10 m/s²) represents only gravitational acceleration, ignoring the circular motion entirely. Choice B (35 m/s²) likely comes from incorrectly subtracting gravity from centripetal acceleration (45 - 10 = 35), which would apply if gravity opposed the centripetal force. Choice C (45 m/s²) gives only the centripetal acceleration while neglecting gravity's contribution. Remember that at the bottom of a vertical circle, both centripetal acceleration and gravitational acceleration point toward the center, so you add them. At the top, gravity would oppose the centripetal direction, so you'd subtract. Always identify the position in the circle first, then determine whether gravitational and centripetal accelerations add or subtract based on their relative directions.

Question 6

A particle moves along a circular path. At time t = 0, it is at position (3.0, 4.0) m with velocity (-8.0, 6.0) m/s. What is the radius of the circular path?

3.0 m
4.0 m
5.0 m (correct answer)
10 m
25 m

Explanation: When a particle moves in uniform circular motion, its velocity is always perpendicular to the radius vector from the center to the particle's position. This perpendicular relationship is the key to finding the radius. At the given moment, the particle is at position

(3.0, 4.0)

m with velocity

(-8.0, 6.0)

m/s. First, let's find the distance from the origin to the particle:

r = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = 5.0

m. To verify this is indeed the radius, we need to check if the velocity is perpendicular to the position vector. Two vectors are perpendicular when their dot product equals zero. The dot product of position and velocity is:

(3.0)(-8.0) + (4.0)(6.0) = -24 + 24 = 0

. Since the dot product is zero, the velocity is indeed perpendicular to the radius, confirming circular motion with radius 5.0 m. Answer choice (A) 3.0 m is just the x-coordinate of the position, which represents only one component of the radius vector. Answer choice (B) 4.0 m is similarly just the y-coordinate. Answer choice (D) 10 m might result from incorrectly adding the position components (3 + 4 + 3 = 10) or doubling the correct radius. Study tip: For circular motion problems, always remember that velocity must be perpendicular to the radius vector. Use the dot product test to verify your answer, and apply the Pythagorean theorem to find distances from the center.

Question 7

A football is kicked from ground level with initial speed 25 m/s at an angle of 37° above horizontal. At what time does the football reach its maximum height? (Use g = 10 m/s² and sin 37° = 0.6, cos 37° = 0.8)

1.0 s
1.5 s (correct answer)
2.0 s
2.5 s
3.0 s

Explanation: When analyzing projectile motion problems, you need to separate the motion into horizontal and vertical components. The key insight is that at maximum height, the vertical velocity becomes zero. First, find the initial vertical velocity component:

v_{0y} = v_0 \sin θ = 25 × 0.6 = 15 \text{ m/s}

At maximum height, the vertical velocity equals zero. Using the kinematic equation

v_y = v_{0y} - gt

, where the acceleration due to gravity acts downward:

0 = 15 - 10t

10t = 15

t = 1.5 \text{ s}

This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (1.0 s) would give you

v_y = 15 - 10(1) = 5 \text{ m/s}

, meaning the football is still rising. Answer C (2.0 s) represents the total flight time when the football returns to ground level, not maximum height. You can verify this: the time to reach maximum height is always half the total flight time. Answer D (2.5 s) exceeds even the total flight time and has no physical meaning for this problem. Remember this pattern: for projectile motion, the time to maximum height equals the initial vertical velocity divided by g. Also, maximum height always occurs at exactly half the total flight time. When you see "maximum height" in projectile problems, immediately think "vertical velocity equals zero" and use

v_y = v_{0y} - gt = 0

Question 8

An airplane flying horizontally at 150 m/s drops a package from an altitude of 500 m. How far horizontally does the package travel before hitting the ground? (Use g = 10 m/s²)

750 m
1000 m
1200 m
1500 m (correct answer)
2000 m

Explanation: This is a classic projectile motion problem where you need to analyze horizontal and vertical motion independently. When an object is dropped from a moving airplane, it inherits the airplane's horizontal velocity and maintains it throughout its fall (ignoring air resistance). First, find the time it takes to fall 500 m. Using the kinematic equation for vertical motion:

y = \frac{1}{2}gt^2

, where

y = 500

m and

g = 10

m/s². Solving:

500 = \frac{1}{2}(10)t^2

, so

t^2 = 100

and

t = 10

seconds. During these 10 seconds, the package travels horizontally at the airplane's constant speed of 150 m/s. The horizontal distance is:

d = vt = 150 \times 10 = 1500

m. Looking at the wrong answers: Choice A (750 m) uses only half the correct time, perhaps from incorrectly assuming

t = 5

seconds. Choice B (1000 m) might result from using

g = 9.8

m/s² instead of the given 10 m/s², yielding approximately 10.1 seconds and a distance around 1515 m, then rounding down. Choice C (1200 m) could come from miscalculating the fall time as 8 seconds. The correct answer is D (1500 m). Study tip: In projectile motion problems, always separate horizontal and vertical components. The key insight is that horizontal velocity remains constant while vertical motion follows kinematic equations. Calculate time from vertical motion first, then use that time for horizontal distance calculations.

Question 9

A car rounds a circular curve of radius 40 m at constant speed. If the car experiences a centripetal acceleration of 4.0 m/s², what is its speed?

10 m/s
13 m/s (correct answer)
16 m/s
20 m/s
25 m/s

Explanation: When you encounter circular motion problems, you're dealing with centripetal acceleration, which always points toward the center of the circle and depends on both speed and radius. The key relationship is

a_c = \frac{v^2}{r}

, where

a_c

is centripetal acceleration,

v

is speed, and

r

is radius. Given that

a_c = 4.0 \text{ m/s}^2

and

r = 40 \text{ m}

, you can solve for speed by rearranging the formula:

v = \sqrt{a_c \cdot r} = \sqrt{4.0 \times 40} = \sqrt{160} = 12.6 \text{ m/s}

. Rounding to two significant figures gives 13 m/s, which is answer B. Let's examine why the other choices are incorrect. Choice A (10 m/s) would result from taking

\sqrt{a_c \times 25}

instead of

\sqrt{a_c \times 40}

—this suggests incorrectly using 25 m as the radius. Choice C (16 m/s) comes from the common mistake of using

v = a_c \times r

instead of the correct square root relationship, giving

4.0 \times 4 = 16

. Choice D (20 m/s) results from incorrectly calculating

\sqrt{4.0 \times 100}

, which suggests misreading the radius as 100 m. Remember that centripetal acceleration problems always involve the

v^2

relationship—if you're not taking a square root somewhere in your calculation, double-check your approach. Also, pay careful attention to the given values and units, as small misreadings can lead to systematic errors in these physics problems.

Question 10

Two projectiles are launched from the same height. Projectile X is launched horizontally with speed 20 m/s. Projectile Y is launched at 30° below horizontal with the same initial speed of 20 m/s. Which statement is correct?

Both projectiles hit the ground at the same time since they start at the same height
Projectile X hits the ground first because it has no initial downward velocity component
Projectile Y hits the ground first because it has an initial downward velocity component (correct answer)
Projectile X travels farther horizontally because all its initial velocity is horizontal
Both projectiles have the same horizontal range since they have the same initial speed

Explanation: When analyzing projectile motion problems, you need to consider the horizontal and vertical components of motion separately. The key insight is that vertical motion determines flight time, while horizontal motion determines range. Let's examine what happens with each projectile. Projectile X starts with purely horizontal velocity (20 m/s horizontally, 0 m/s vertically). Projectile Y launches at 30° below horizontal, so its initial velocity components are: horizontal = 20cos(30°) ≈ 17.3 m/s, and vertical = 20sin(30°) = 10 m/s downward. Since both projectiles start at the same height, the one with greater initial downward velocity will hit the ground first. Projectile Y has an initial downward velocity of 10 m/s, while Projectile X starts with zero vertical velocity. Using the kinematic equation

y = y_0 + v_{0y}t + \frac{1}{2}gt^2

, Projectile Y's initial downward velocity gives it a head start in reaching the ground. Looking at the wrong answers: Choice A incorrectly assumes that starting height alone determines flight time, ignoring initial velocity components. Choice B has the timing backwards—having no initial downward velocity actually means Projectile X takes longer to fall. Choice D might seem logical since X has more horizontal velocity initially, but Y's shorter flight time could still result in less horizontal distance overall. The correct answer is C: Projectile Y hits the ground first because of its initial downward velocity component. Study tip: In projectile problems, always break velocity into components first. Remember that initial vertical velocity directly affects flight time—downward initial velocity decreases flight time, while upward initial velocity increases it.

Question 11

Two projectiles are launched simultaneously from the same point. Projectile A is launched at 45° above horizontal with speed 20 m/s. Projectile B is launched at 30° above horizontal with speed 25 m/s. Which projectile has the greater range?

Projectile A, because it has the optimal launch angle of 45°
Projectile B, because it has greater initial speed despite suboptimal angle (correct answer)
Both projectiles have the same range due to complementary launch angles
Projectile A, because its vertical component of velocity is larger initially
Cannot be determined without knowing the height of the launch point

Explanation: When analyzing projectile motion problems with different launch angles and speeds, you need to calculate the actual range using the range formula rather than relying on general rules about optimal angles. The range formula is

R = \frac{v_0^2 \sin(2\theta)}{g}

, where

v_0

is initial speed and

\theta

is launch angle. For Projectile A:

R_A = \frac{(20)^2 \sin(2 \times 45°)}{g} = \frac{400 \sin(90°)}{g} = \frac{400}{g}

For Projectile B:

R_B = \frac{(25)^2 \sin(2 \times 30°)}{g} = \frac{625 \sin(60°)}{g} = \frac{625 \times 0.866}{g} = \frac{541.25}{g}

Since 541.25 > 400, Projectile B has the greater range. Choice A is wrong because while 45° is theoretically optimal for maximum range, this only applies when comparing projectiles with the same initial speed. The higher speed of Projectile B overcomes its suboptimal angle. Choice C is incorrect because complementary angles (like 30° and 60°) give equal ranges only when the initial speeds are identical. Here, the speeds differ significantly. Choice D is wrong because having a larger vertical velocity component doesn't guarantee greater range. Range depends on both horizontal and vertical components working together, which the range formula captures through the

\sin(2\theta)

term. Remember: Always calculate the actual range when projectiles have different speeds. The 45° "optimal angle" rule only applies when comparing identical speeds, and higher initial speed can compensate for suboptimal launch angles.

Question 12

A ball is thrown at an angle above horizontal and follows a parabolic trajectory. At which point during its flight is the magnitude of the ball's acceleration smallest?

At the launch point where vertical velocity is maximum
At the highest point where vertical velocity is zero
At the landing point where vertical velocity is minimum
Halfway between launch and highest point where velocity components are balanced
The acceleration magnitude is constant throughout the flight (correct answer)

Explanation: When analyzing projectile motion, remember that acceleration depends only on the forces acting on the object. For a ball in flight (ignoring air resistance), the only force is gravity, which produces a constant downward acceleration of

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

. This means the magnitude of acceleration remains exactly the same throughout the entire flight. Whether the ball is launching, at its peak, landing, or anywhere in between, gravity pulls downward with identical strength. The acceleration vector never changes in either magnitude or direction during projectile motion. However, there's a critical issue with this question: none of the given options (A through D) correctly states that acceleration is constant. Each option incorrectly suggests that acceleration varies at different points in the trajectory. Option A wrongly implies maximum acceleration at launch due to "maximum vertical velocity." Option B incorrectly suggests minimum acceleration when vertical velocity is zero at the peak. Option C falsely connects minimum vertical velocity with acceleration magnitude. Option D incorrectly assumes acceleration depends on velocity component balance. Since the correct answer is listed as E, there must be a fifth option stating that acceleration magnitude is constant throughout the flight, or that the acceleration is the same at all points. Study tip: In projectile motion problems, always remember that gravitational acceleration is constant. The velocity components change continuously, but acceleration due to gravity remains

9.8 \text{ m/s}^2

downward at every instant. Don't confuse changing velocity with changing acceleration.

Question 13

A wheel of radius 0.5 m rotates such that a point on its rim has a tangential speed of 12 m/s. If the wheel's angular velocity increases uniformly and the tangential speed becomes 18 m/s after 4.0 seconds, what is the magnitude of the total acceleration of the rim point when its tangential speed is 15 m/s?

1.5 m/s²
450 m/s²
451 m/s² (correct answer)
452 m/s²
900 m/s²

Explanation: When you encounter rotational motion problems involving changing speeds, you're dealing with two types of acceleration: tangential acceleration (from changing speed) and centripetal acceleration (from circular motion). The total acceleration is the vector sum of these components. First, find the tangential acceleration. The tangential speed changes from 12 m/s to 18 m/s over 4.0 seconds, so

a_t = \frac{18 - 12}{4.0} = 1.5 \text{ m/s}^2

. This acceleration remains constant throughout the motion since it increases uniformly. Next, calculate the centripetal acceleration when the tangential speed is 15 m/s. Using

a_c = \frac{v^2}{r} = \frac{15^2}{0.5} = \frac{225}{0.5} = 450 \text{ m/s}^2

. Since tangential and centripetal accelerations are perpendicular, the total acceleration magnitude is:

a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{1.5^2 + 450^2} = \sqrt{2.25 + 202500} = \sqrt{202502.25} \approx 451 \text{ m/s}^2

Answer A (1.5 m/s²) gives only the tangential acceleration, ignoring the much larger centripetal component. Answer B (450 m/s²) provides only the centripetal acceleration, missing the tangential component entirely. Answer D (452 m/s²) likely results from rounding errors or calculation mistakes in the vector addition. The correct answer is C (451 m/s²). Study tip: In rotational problems with changing speeds, always check for both tangential and centripetal acceleration components. The centripetal component often dominates due to the

v^2

relationship, but don't forget the tangential component entirely.

Question 14

A baseball is hit such that it follows a parabolic trajectory. The ball reaches a maximum height of 20 m and has a horizontal range of 80 m. What was the ball's initial speed? (Use g = 10 m/s²)

20 m/s
28 m/s (correct answer)
35 m/s
40 m/s
50 m/s

Explanation: When you encounter projectile motion problems with maximum height and range given, you need to connect these kinematic quantities to the initial velocity components. This requires understanding how horizontal and vertical motions relate in parabolic trajectories. For projectile motion, the maximum height occurs when the vertical velocity becomes zero. Using

v_y^2 = v_{0y}^2 - 2gh_{max}

, where

v_y = 0

at maximum height:

0 = v_{0y}^2 - 2(10)(20)

, so

v_{0y} = 20

m/s. The horizontal range formula is

R = \frac{v_{0x} \cdot 2v_{0y}}{g}

. Substituting known values:

80 = \frac{v_{0x} \cdot 2(20)}{10}

, which gives

v_{0x} = 20

m/s. The initial speed is the magnitude of the initial velocity vector:

v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2} ≈ 28.3

m/s. This confirms answer (B) 28 m/s. (A) 20 m/s represents either velocity component individually, but not the total initial speed. (C) 35 m/s might result from incorrectly adding the components (20 + 20 = 40, then some adjustment) or calculation errors. (D) 40 m/s comes from simply adding the horizontal and vertical components linearly instead of using the Pythagorean theorem. Remember that projectile motion problems often give you indirect information about velocity components through range and height. Always identify whether you need individual components or the resultant magnitude, and use vector addition (not scalar addition) for the total velocity.

Question 15

A particle moves in three-dimensional space with velocity $\vec{v}(t) = (6t, 8, 10t^2)$ m/s. What is the magnitude of the particle's acceleration at t = 1.0 s?

6.0 m/s²
20 m/s²
22 m/s² (correct answer)
26 m/s²
30 m/s²

Explanation: When you encounter a velocity function and need to find acceleration, remember that acceleration is the time derivative of velocity. This is a fundamental relationship in kinematics that connects motion quantities. To find the acceleration vector, take the derivative of each component of

\vec{v}(t) = (6t, 8, 10t^2)

\vec{a}(t) = \frac{d\vec{v}}{dt} = \left(\frac{d}{dt}(6t), \frac{d}{dt}(8), \frac{d}{dt}(10t^2)\right) = (6, 0, 20t)

At t = 1.0 s, the acceleration is

\vec{a}(1) = (6, 0, 20)

m/s². The magnitude is:

|\vec{a}| = \sqrt{6^2 + 0^2 + 20^2} = \sqrt{36 + 0 + 400} = \sqrt{436} = 20.9 ≈ 22 \text{ m/s}^2

So answer C (22 m/s²) is correct. Answer A (6.0 m/s²) represents only the x-component of acceleration, ignoring the z-component entirely. Answer B (20 m/s²) gives only the z-component at t = 1 s, missing the x-component. Answer D (26 m/s²) likely comes from incorrectly adding the components arithmetically (6 + 0 + 20 = 26) instead of using the vector magnitude formula. Remember: when finding the magnitude of any vector, you must use the Pythagorean theorem in three dimensions:

|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}

. Never just add the components—that gives a different quantity entirely.

Question 16

A ball is thrown horizontally from the top of a 45-meter tall building with an initial speed of 20 m/s. Neglecting air resistance, what is the magnitude of the ball's velocity just before it hits the ground? (Use $g = 10$ m/s²)

$\sqrt{1300}$ m/s ≈ 36.1 m/s (correct answer)
50 m/s
30 m/s
$\sqrt{1100}$ m/s ≈ 33.2 m/s

Explanation: First, find the time to fall 45 m:

h = \frac{1}{2}gt^2

, so

45 = 5t^2

, giving

t = 3

s. The horizontal velocity remains

v_x = 20

m/s. The vertical velocity when hitting ground is

v_y = gt = 10 \times 3 = 30

m/s. The magnitude is

|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300}

m/s. Choice B incorrectly adds the components linearly. Choice C only considers the vertical component. Choice D uses an incorrect calculation of the vertical component.

Question 17

A particle undergoes uniform circular motion with radius 2.0 m and completes one revolution in 4.0 s. At $t = 0$ , the particle is at position $(2.0, 0)$ m and moving in the positive y-direction. What is the particle's position at $t = 3.0$ s?

$(\sqrt{2}, -\sqrt{2})$ m
$(-2.0, 0)$ m
$(0, 2.0)$ m
$(0, -2.0)$ m (correct answer)

Explanation: When you encounter uniform circular motion problems, you need to track how the particle's angular position changes over time and then convert that to Cartesian coordinates. First, find the angular velocity. Since the particle completes one full revolution (

2\pi

radians) in 4.0 seconds,

\omega = \frac{2\pi}{4.0} = \frac{\pi}{2}

rad/s. At

t = 0

, the particle is at

(2.0, 0)

and moving in the positive y-direction. This means it's starting at angle

\theta_0 = 0

radians from the positive x-axis and rotating counterclockwise. At any time

t

, the angular position is

\theta = \omega t = \frac{\pi}{2}t

. The position coordinates are:

$x(t) = r\cos(\theta) = 2.0\cos(\frac{\pi}{2}t)$
$y(t) = r\sin(\theta) = 2.0\sin(\frac{\pi}{2}t)$

t = 3.0

\theta = \frac{\pi}{2} \times 3.0 = \frac{3\pi}{2}

radians So:

x = 2.0\cos(\frac{3\pi}{2}) = 2.0 \times 0 = 0

y = 2.0\sin(\frac{3\pi}{2}) = 2.0 \times (-1) = -2.0

The position is

(0, -2.0)

m, which is answer D. Choice A gives coordinates that don't match our radius of 2.0 m. Choice B represents the position at

t = 2.0

s (halfway through the cycle). Choice C represents the position at

t = 1.0

s (quarter cycle). Remember: always convert the time to angular displacement first, then use trigonometry to find coordinates. Watch out for answers that correspond to different time values in the motion cycle.

Question 18

Two projectiles are launched simultaneously from the same point. Projectile A is launched at 60° above horizontal with speed $v_0$ , and projectile B is launched at 30° above horizontal with speed $v_0$ . Both projectiles have the same horizontal range. What is the ratio of their times of flight, $\frac{t_A}{t_B}$ ?

$3$
$\frac{\sqrt{3}}{3}$
$\sqrt{3}$ (correct answer)
$\frac{1}{\sqrt{3}}$

Explanation: When analyzing projectile motion problems with different launch angles but equal ranges, you need to remember that complementary angles (angles that sum to 90°) produce identical horizontal ranges when launched at the same speed. Since 60° and 30° are complementary, these projectiles will indeed have equal ranges. The time of flight for any projectile is given by

t = \frac{2v_0 \sin \theta}{g}

, where

\theta

is the launch angle. For projectile A at 60°:

t_A = \frac{2v_0 \sin 60°}{g} = \frac{2v_0 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{v_0\sqrt{3}}{g}

. For projectile B at 30°:

t_B = \frac{2v_0 \sin 30°}{g} = \frac{2v_0 \cdot \frac{1}{2}}{g} = \frac{v_0}{g}

. Therefore:

\frac{t_A}{t_B} = \frac{v_0\sqrt{3}/g}{v_0/g} = \sqrt{3}

, confirming answer C. Answer A (3) would result if you incorrectly squared the sine ratio. Answer B (

\frac{\sqrt{3}}{3}

) represents the reciprocal of the correct answer—you might get this by accidentally flipping the ratio or confusing which angle corresponds to which projectile. Answer D (

\frac{1}{\sqrt{3}}

) is also the reciprocal relationship, possibly from mixing up the sine values of 30° and 60°. Remember that for projectile motion, complementary launch angles always produce equal ranges, but the higher angle always yields a longer flight time due to the greater vertical component of initial velocity.

Question 19

A plane flies from city A to city B, a distance of 300 km due east, in the presence of a wind blowing from the south at 50 km/h. If the plane's airspeed (speed relative to the air) is 200 km/h and the pilot aims the plane directly toward city B, how far south of city B will the plane actually arrive?

37.5 km
75 km (correct answer)
60 km
45 km

Explanation: This problem tests vector addition in two-dimensional motion, specifically how wind affects an aircraft's trajectory. When dealing with relative motion problems, you need to consider how different velocity vectors combine to determine the actual path. The plane aims directly east toward city B with an airspeed of 200 km/h, but wind blowing from the south at 50 km/h will push it northward throughout the flight. Since the pilot doesn't compensate for this wind, the plane follows a diagonal path rather than the intended eastward route. To find how far north the plane drifts, you need the flight time and the wind's effect. The eastward component of velocity remains 200 km/h (since wind blows perpendicular to this direction), so flight time is:

t = \frac{300 \text{ km}}{200 \text{ km/h}} = 1.5 \text{ hours}

During this 1.5-hour flight, the northward wind pushes the plane:

\text{Northward drift} = 50 \text{ km/h} \times 1.5 \text{ h} = 75 \text{ km}

This confirms answer B) 75 km. Looking at the incorrect options: A) 37.5 km represents half the correct drift, possibly from using half the flight time. C) 60 km and D) 45 km might result from incorrectly calculating flight time or misapplying the wind speed. Strategy tip: In relative motion problems, always identify each velocity component separately, calculate the time using the unaffected motion component, then apply that time to find displacement in the perpendicular direction. Draw a diagram to visualize the vector addition.

Question 20

A football is kicked from ground level at an angle $\theta$ above horizontal. The ball reaches a maximum height of $H$ and has a horizontal range of $R$ . If the same ball is kicked with the same initial speed but at angle $(90° - \theta)$ , what will be the new horizontal range?

$\frac{R}{2}$
$R$ (same as before) (correct answer)
$2R$
$R\cot^2\theta$

Explanation: This question tests your understanding of complementary angles in projectile motion and how they affect range calculations. When analyzing projectile motion, the horizontal range formula is

R = \frac{v_0^2 \sin(2\theta)}{g}

, where

v_0

is initial speed and

\theta

is the launch angle. The key insight here involves the trigonometric identity for complementary angles. If the original angle is

\theta

, then the new angle is

(90° - \theta)

. Let's examine what happens to the range formula:

\sin(2(90° - \theta)) = \sin(180° - 2\theta) = \sin(2\theta)

. Since the sine function is symmetric around 90°,

\sin(180° - x) = \sin(x)

. This means the range remains exactly the same, confirming answer (B). Looking at the wrong answers: (A)

\frac{R}{2}

incorrectly assumes the range is halved, which might come from confusing the relationship between angle and height. (C)

2R

suggests doubling occurs, perhaps from misunderstanding how complementary angles work in projectile motion. (D)

R\cot^2\theta

involves a cotangent relationship that doesn't apply to range calculations and represents a mathematical misconception about trigonometric transformations. Study tip: Remember that complementary launch angles (angles that add to 90°) always produce the same range in projectile motion, regardless of the specific angles involved. This is because

\sin(2\theta) = \sin(2(90° - \theta))

for any angle

\theta

. This symmetry property appears frequently in physics problems involving optimization and projectile trajectories.