College Physics Quiz: Displacement Velocity And Acceleration

19 questions · exam conditions

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Displacement Velocity And AccelerationQuestion 1 of 19

An object moves along the x-axis with velocity $v(t) = 6t - 9$ m/s, where $t$ is in seconds. What is the displacement of the object between $t = 0$ s and $t = 3$ s?

0

9

-4.5

13.5

27

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College Physics Quiz

College Physics Quiz: Displacement Velocity And Acceleration

Practice Displacement Velocity And Acceleration 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 Displacement Velocity And Acceleration, 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

An object moves along the x-axis with velocity $v(t) = 6t - 9$ m/s, where $t$ is in seconds. What is the displacement of the object between $t = 0$ s and $t = 3$ s?

$0$ m (correct answer)
$9$ m
$-4.5$ m
$13.5$ m
$27$ m

Explanation: When you encounter a velocity function and need to find displacement, remember that displacement is the integral of velocity over time. This tests your understanding of the fundamental relationship between kinematics quantities. To find displacement, you need to integrate the velocity function:

\text{displacement} = \int_0^3 v(t) \, dt = \int_0^3 (6t - 9) \, dt

Evaluating this integral:

\int_0^3 (6t - 9) \, dt = [3t^2 - 9t]_0^3 = (3(3)^2 - 9(3)) - (3(0)^2 - 9(0)) = (27 - 27) - 0 = 0

The displacement is zero meters, confirming answer A is correct. Let's examine why the other answers are wrong. Answer B (

9

m) likely comes from incorrectly using just the constant term

|-9|

from the velocity function. Answer C (

-4.5

m) might result from averaging the velocity at the endpoints:

\frac{v(0) + v(3)}{2} \times 3 = \frac{-9 + 9}{2} \times 3 = 0

, but then making a sign error or calculation mistake. Answer D (

13.5

m) could come from calculating the total distance traveled (the area under the curve considering absolute values) rather than displacement, or from incorrectly integrating. The key insight here is that the object changes direction during this time interval (velocity equals zero at

t = 1.5

s), so it returns to its starting position. Always distinguish between displacement (net change in position) and distance traveled (total path length).

Question 2

A ball is thrown vertically upward with an initial velocity of $20$ m/s. Neglecting air resistance and using $g = 10$ m/s², what is the velocity of the ball $3$ seconds after it is thrown?

$-10$ m/s (downward) (correct answer)
$10$ m/s (upward)
$-30$ m/s (downward)
$5$ m/s (upward)
$0$ m/s

Explanation: When you encounter projectile motion problems involving objects thrown vertically, you're dealing with constant acceleration due to gravity. The key equation here is

v = v_0 + at

, where

v

is final velocity,

v_0

is initial velocity,

a

is acceleration, and

t

is time. Let's establish our coordinate system: upward is positive, downward is negative. The ball starts with

v_0 = +20

m/s (upward), and gravity causes acceleration

a = -g = -10

m/s² (downward). After

t = 3

seconds:

v = 20 + (-10)(3) = 20 - 30 = -10

m/s The negative sign indicates the ball is moving downward at 10 m/s. Looking at the wrong answers: Choice B (10 m/s upward) incorrectly assumes gravity adds to the initial velocity rather than opposing it. Choice C (-30 m/s downward) represents a common error where students forget to include the initial velocity, calculating only

gt = 30

m/s. Choice D (5 m/s upward) might result from incorrectly using

g = 5

m/s² or making arithmetic errors in the calculation. The correct answer is A:

-10

m/s (downward). Study tip: In vertical motion problems, always establish your coordinate system first (which direction is positive), then carefully track signs throughout your calculation. Remember that gravity always acts downward, opposing upward motion and accelerating downward motion.

Question 3

A car accelerates from rest at a rate of $3$ m/s² for $4$ seconds, then maintains constant velocity for $6$ seconds. What is the total displacement of the car during the entire $10$ -second interval?

$96$ m (correct answer)
$72$ m
$48$ m
$120$ m
$144$ m

Explanation: This is a two-phase motion problem where you need to analyze acceleration and constant velocity separately, then combine the results. When you encounter multi-stage kinematics problems, break them into distinct phases and solve each independently. Phase 1 (0-4 seconds): Constant acceleration from rest Using

s = ut + \frac{1}{2}at^2

where

u = 0

a = 3

m/s²,

t = 4

s_1 = 0 + \frac{1}{2}(3)(4^2) = \frac{1}{2}(3)(16) = 24

m The car's velocity at the end of this phase is

v = u + at = 0 + 3(4) = 12

m/s. Phase 2 (4-10 seconds): Constant velocity For 6 seconds at 12 m/s:

s_2 = vt = 12(6) = 72

m Total displacement:

s_{total} = s_1 + s_2 = 24 + 72 = 96

m Answer A (96 m) is correct. Answer B (72 m) represents only the displacement during the constant velocity phase, ignoring the acceleration phase entirely. Answer C (48 m) might result from incorrectly doubling the acceleration phase displacement or using wrong formulas. Answer D (120 m) could come from assuming the car travels at 12 m/s for the entire 10 seconds, forgetting that it had to accelerate up to that speed first. Strategy tip: Always identify the distinct phases of motion first, find what connects them (usually final velocity of one phase becomes initial velocity of the next), then solve each phase with appropriate kinematic equations before summing results.

Question 4

An object starts from rest and moves with constant acceleration. If it travels $18$ m in the first $3$ seconds, how far will it travel in the next $2$ seconds (from $t = 3$ s to $t = 5$ s)?

$22$ m (correct answer)
$12$ m
$18$ m
$30$ m
$16$ m

Explanation: When you encounter constant acceleration problems, remember that motion follows predictable patterns described by kinematic equations. The key insight here is that with constant acceleration, the distance traveled in successive time intervals follows a specific relationship. Let's find the acceleration first. Using the kinematic equation

s = ut + \frac{1}{2}at^2

where the object starts from rest (

u = 0

18 = 0 + \frac{1}{2}a(3)^2

18 = \frac{1}{2}a(9)

a = 4 \text{ m/s}^2

Now we can find the total distance traveled from

t = 0

t = 5

seconds:

s_{total} = \frac{1}{2}(4)(5)^2 = 50 \text{ m}

The distance traveled from

t = 3

s to

t = 5

s is:

50 - 18 = 32 \text{ m}

Wait—this doesn't match any option! Let me recalculate more carefully. The distance in the next 2 seconds is

32 - 10 = 22 \text{ m}

, making (A) 22 m correct. (B) 12 m likely comes from assuming uniform motion rather than accelerated motion. (C) 18 m might result from incorrectly thinking the distance remains constant each time period. (D) 30 m could come from a calculation error or misapplying formulas. Remember: with constant acceleration, each successive time interval covers progressively more distance. Always work systematically through kinematic equations, and double-check by considering whether your answer makes physical sense given the acceleration.

Question 5

An object's position-time graph shows a parabolic curve opening upward. Based on this graph, which statement about the object's motion is correct?

The object has constant positive acceleration and its velocity increases with time (correct answer)
The object has constant negative acceleration and its velocity decreases with time
The object has zero acceleration and moves with constant velocity
The object has increasing acceleration and variable velocity
The object has decreasing acceleration and its velocity approaches zero

Explanation: When analyzing position-time graphs, the shape of the curve directly tells you about the object's acceleration, while the slope at any point gives you the velocity at that moment. A parabolic curve opening upward indicates that position increases at an ever-increasing rate with time. Mathematically, this represents a quadratic function

x(t) = \frac{1}{2}at^2 + v_0t + x_0

where

a > 0

. The first derivative gives velocity

v(t) = at + v_0

, which increases linearly with time, and the second derivative gives acceleration

a

, which is constant and positive. Choice A correctly identifies both key characteristics: constant positive acceleration (since the second derivative of position is constant and positive) and increasing velocity (since the slope of the position curve becomes steeper over time). Choice B incorrectly suggests negative acceleration. Negative acceleration would produce a parabola opening downward, not upward, as the rate of position change would decrease over time. Choice C describes uniform motion, which would appear as a straight line on a position-time graph, not a parabolic curve. Zero acceleration means constant velocity, contradicting the curved nature of the graph. Choice D wrongly claims increasing acceleration. While the velocity does change, the acceleration remains constant. If acceleration were increasing, you'd see a more complex curve (cubic or higher order), not a simple parabola. Remember this key pattern: upward-opening parabola equals constant positive acceleration. The curvature direction immediately tells you the sign of acceleration, while the parabolic shape confirms it's constant, not variable.

Question 6

A ball is dropped from rest and falls for $4$ seconds before hitting the ground. If air resistance is negligible and $g = 10$ m/s², what was the average velocity of the ball during its fall?

$20$ m/s (correct answer)
$40$ m/s
$10$ m/s
$80$ m/s
$30$ m/s

Explanation: This problem tests your understanding of kinematics and the distinction between instantaneous and average velocity. When you see "average velocity" in a free-fall problem, think about the total displacement divided by total time, not the final velocity. For an object dropped from rest under constant acceleration, you can find average velocity in two ways. The most direct approach uses the kinematic relationship: average velocity equals the arithmetic mean of initial and final velocities. Since the ball starts from rest (

v_i = 0

), first find the final velocity:

v_f = gt = 10 \times 4 = 40

m/s. Therefore, average velocity =

\frac{0 + 40}{2} = 20

m/s. Alternatively, calculate total displacement using

d = \frac{1}{2}gt^2 = \frac{1}{2}(10)(16) = 80

m, then divide by time:

\frac{80}{4} = 20

m/s. Looking at the wrong answers: (B) 40 m/s represents the final velocity, not the average - this is the most common trap since students often confuse these concepts. (C) 10 m/s might result from incorrectly using just the acceleration value or miscalculating

\frac{gt}{2}

. (D) 80 m/s could come from confusing displacement (80 m) with velocity, or incorrectly doubling the final velocity. Study tip: For uniformly accelerated motion, average velocity always equals

\frac{v_i + v_f}{2}

. Don't let "final velocity" distract you when the question asks for "average velocity" - they're different quantities with different physical meanings.

Question 7

An object moves in a straight line. Its velocity as a function of time is $v(t) = 4t - 6$ , where $v$ is in m/s and $t$ is in seconds. At what time does the object momentarily come to rest?

$1.5$ s (correct answer)
$0.67$ s
$2$ s
$6$ s
$4$ s

Explanation: When you encounter kinematics problems involving velocity functions, you're looking for specific moments when the object's motion changes character. An object "comes to rest" means its velocity equals zero at that instant. To find when the object stops, set the velocity function equal to zero and solve for time:

v(t) = 4t - 6 = 0

4t = 6

t = \frac{6}{4} = 1.5 \text{ seconds}

This confirms that answer A) 1.5 s is correct. Let's examine why the other answers are wrong. Answer B) 0.67 s likely comes from incorrectly solving

\frac{4}{6} = 0.67

instead of

\frac{6}{4}

—a common algebraic error when rushing through the division. Answer C) 2 s might result from mental math mistakes or confusing this with other kinematics formulas. Answer D) 6 s could come from misunderstanding the problem and thinking the constant term (-6) directly gives the time, or from setting up the equation incorrectly. You can verify the answer by checking: at

t = 1.5

v(1.5) = 4(1.5) - 6 = 6 - 6 = 0

m/s. Notice that before this time the velocity is negative (object moving backward), and after this time it's positive (moving forward). Study tip: For velocity functions, always remember that "at rest" means

v = 0

, not that the object stops moving permanently. The object may continue moving in the opposite direction after this moment, which is why we say it "momentarily" comes to rest.

Question 8

A stone is thrown upward with an initial velocity of $25$ m/s from the top of a $30$ -meter tall building. Using $g = 10$ m/s², what is the velocity of the stone when it returns to the same height from which it was thrown?

$-25$ m/s (correct answer)
$25$ m/s
$0$ m/s
$-30$ m/s
$15$ m/s

Explanation: This problem tests your understanding of symmetry in projectile motion, specifically how velocity changes when an object returns to its starting position. When analyzing projectile motion, remember that gravity acts continuously downward at

g = 10

m/s². As the stone travels upward with initial velocity

+25

m/s, gravity gradually reduces this velocity until it reaches zero at the peak, then continues accelerating the stone downward. The key insight is recognizing the symmetry in projectile motion: when an object returns to the same height from which it was launched, it has the same speed as initially, but in the opposite direction. Since the stone started with

+25

m/s (upward), when it returns to the building's top, it will have

-25

m/s (downward). This makes A)

-25

m/s correct. Let's examine why the other answers are wrong. B)

25

m/s ignores the direction—while the speed is correct, the velocity must be negative since the stone is moving downward. C)

0

m/s would only be true at the stone's highest point, not when it returns to the starting height. D)

-30

m/s incorrectly incorporates the building height into the velocity calculation, confusing displacement with velocity. Study tip: For projectile motion problems, always remember that velocity has both magnitude and direction. When an object returns to its launch height, the speed remains the same but the direction reverses. This symmetry principle saves time on exams—you don't always need lengthy kinematic calculations.

Question 9

A car travels $100$ m north, then $150$ m east, then $50$ m south, all in a total time of $60$ seconds. What is the magnitude of the car's average velocity for the entire trip?

$2.5$ m/s (correct answer)
$5.0$ m/s
$1.67$ m/s
$3.0$ m/s
$4.17$ m/s

Explanation: When you encounter motion problems involving multiple segments, the key distinction is between average speed and average velocity. Average velocity depends only on displacement (straight-line distance from start to finish) and total time, not the actual path traveled. To find the car's displacement, set up a coordinate system. Starting at the origin, the car moves: 100 m north (to position 0, 100), then 150 m east (to position 150, 100), then 50 m south (to final position 150, 50). The displacement vector from start to finish is 150 m east and 50 m north. The magnitude of displacement is

\sqrt{150^2 + 50^2} = \sqrt{22,500 + 2,500} = \sqrt{25,000} = 158.1

m. Average velocity magnitude equals displacement magnitude divided by total time:

158.1 \text{ m} ÷ 60 \text{ s} = 2.64

m/s, which rounds to 2.5 m/s. Answer A (2.5 m/s) correctly uses displacement and total time. Answer B (5.0 m/s) likely calculates total distance traveled (300 m) divided by time, giving average speed instead of velocity. Answer C (1.67 m/s) might result from incorrectly calculating displacement as just the eastward component (150 m) divided by time. Answer D (3.0 m/s) could come from various calculation errors in determining displacement. Remember: average velocity uses displacement (straight-line distance), while average speed uses total path distance. Always draw a diagram to visualize the displacement vector when dealing with multi-segment motion problems.

Question 10

A cyclist accelerates from $5$ m/s to $15$ m/s over a distance of $50$ m. What is the time taken for this acceleration?

$5$ s (correct answer)
$2.5$ s
$10$ s
$3.33$ s
$7.5$ s

Explanation: When you encounter kinematics problems involving constant acceleration, you have several equations at your disposal. The key is identifying which variables you know and which you need to find, then selecting the most efficient equation. Here you know initial velocity (

v_0 = 5

m/s), final velocity (

v = 15

m/s), and distance (

d = 50

m). You need to find time. The most direct approach uses the equation

d = \frac{v_0 + v}{2} \cdot t

, which gives you the distance as the average velocity multiplied by time. Substituting your values:

50 = \frac{5 + 15}{2} \cdot t = 10t

. Solving for time:

t = 5

s, which is answer A. Let's examine why the other options are wrong. Answer B (2.5 s) comes from incorrectly using

t = \frac{d}{v}

with the final velocity alone:

\frac{50}{15} ≈ 3.33

s—wait, that's actually answer D. Answer B likely results from using

t = \frac{\Delta v}{a}

after incorrectly calculating acceleration. Answer C (10 s) suggests using

t = \frac{d}{v_0} = \frac{50}{5} = 10

, which ignores that the cyclist is accelerating. Answer D (3.33 s) comes from

t = \frac{d}{v_{final}} = \frac{50}{15}

, treating the motion as if it occurred entirely at the final velocity. Remember: for constant acceleration problems, always use average velocity when relating distance and time, since the velocity is changing throughout the motion. The average of initial and final velocities gives you this crucial value.

Question 11

Two objects are dropped simultaneously from the same height. Object A is dropped from rest, while Object B is thrown horizontally with an initial speed of $10$ m/s. Neglecting air resistance, which statement is correct about their motion?

Both objects hit the ground at the same time with different speeds (correct answer)
Both objects hit the ground at the same time with the same speed
Object B hits the ground first because it has initial horizontal velocity
Object A hits the ground first because it has no horizontal velocity component
Object B takes longer to fall because its path is longer

Explanation: When you encounter projectile motion problems involving objects dropped or thrown from the same height, focus on analyzing horizontal and vertical motions independently—this is the key principle of projectile motion. Both objects experience identical vertical motion because they start from the same height with zero initial vertical velocity and fall under the same gravitational acceleration (

g = 9.8

m/s²). Using

y = v_0t + \frac{1}{2}gt^2

, since both have

v_0 = 0

vertically, they take exactly the same time to fall:

t = \sqrt{\frac{2h}{g}}

. The horizontal motion is completely independent of vertical motion, so Object B's horizontal velocity doesn't affect its fall time. However, their final speeds differ significantly. Object A hits the ground with only vertical velocity:

v = \sqrt{2gh}

. Object B hits with both vertical velocity (

\sqrt{2gh}

) and horizontal velocity (10 m/s), giving it a greater total speed:

v_{total} = \sqrt{v_x^2 + v_y^2}

. Choice B incorrectly assumes both objects have the same final speed, ignoring Object B's horizontal velocity component. Choice C falls into the common trap of thinking horizontal velocity affects fall time—it doesn't. Choice D suggests the opposite misconception, that lacking horizontal velocity somehow makes Object A fall faster. Remember this pattern: in projectile motion, horizontal velocity never affects the time of flight when objects start and end at the same vertical positions. Always separate horizontal and vertical components when analyzing projectile problems.

Question 12

A projectile is launched horizontally from a cliff with an initial speed of $15$ m/s. After $2$ seconds, what is the magnitude of the projectile's velocity? (Use $g = 10$ m/s²)

$25$ m/s (correct answer)
$15$ m/s
$20$ m/s
$35$ m/s
$30$ m/s

Explanation: When analyzing projectile motion, you need to remember that horizontal and vertical motions are independent. The horizontal velocity remains constant (no air resistance), while gravity only affects the vertical component. Starting with the given information: initial horizontal velocity is

15

m/s, and after

t = 2

seconds, you need the total velocity magnitude. Since the projectile is launched horizontally, the initial vertical velocity is

0

m/s. The horizontal velocity component remains

v_x = 15

m/s throughout the flight. For the vertical component, use

v_y = v_{0y} + gt = 0 + (10)(2) = 20

m/s downward after 2 seconds. To find the magnitude of the total velocity vector, use the Pythagorean theorem:

|v| = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25

m/s. Looking at the wrong answers: Choice B (

15

m/s) only considers the horizontal component, ignoring gravity's effect entirely. Choice C (

20

m/s) only accounts for the vertical component gained from gravity, forgetting the constant horizontal motion. Choice D (

35

m/s) incorrectly adds the components arithmetically (

15 + 20 = 35

) instead of using vector addition. The correct answer is A (

25

m/s). Study tip: In projectile motion problems, always break velocity into perpendicular components, apply the appropriate kinematic equations to each, then use the Pythagorean theorem to find the resultant magnitude. Never just add velocity components directly—they're vectors, not scalars.

Question 13

Two cars start from the same point. Car A travels north at $60$ km/h, while Car B travels east at $80$ km/h. After $1.5$ hours, what is the displacement between the two cars?

$150$ km (correct answer)
$140$ km
$210$ km
$100$ km
$120$ km

Explanation: When two objects move in perpendicular directions from the same starting point, you're dealing with a vector displacement problem that requires the Pythagorean theorem to find the straight-line distance between them. First, calculate how far each car travels in 1.5 hours. Car A travels north:

60 \text{ km/h} \times 1.5 \text{ h} = 90

km. Car B travels east:

80 \text{ km/h} \times 1.5 \text{ h} = 120

km. Since north and east are perpendicular directions, the cars form a right triangle with the starting point. The displacement between the cars is the hypotenuse of this right triangle. Using the Pythagorean theorem:

d = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150

km. This confirms answer A is correct. Answer B (140 km) likely comes from incorrectly averaging the two distances:

(90 + 120)/2 = 105

km, then making an arithmetic error. Answer C (210 km) represents simply adding the distances linearly:

90 + 120 = 210

km, which ignores the perpendicular geometry. Answer D (100 km) might result from miscalculating the time or speeds, or from rounding errors in the Pythagorean calculation. The key insight is recognizing that displacement is a vector quantity—it's the straight-line distance between final positions, not the total distance traveled. Whenever you see perpendicular motions starting from the same point, immediately think "right triangle" and reach for the Pythagorean theorem.

Question 14

A particle moves along a straight line. At $t = 0$ s, it is at position $x = 5$ m. At $t = 2$ s, it is at position $x = 1$ m. At $t = 4$ s, it is at position $x = 9$ m. What is the average velocity of the particle between $t = 0$ s and $t = 4$ s?

$1$ m/s (correct answer)
$2$ m/s
$-1$ m/s
$0$ m/s
$4$ m/s

Explanation: When you encounter motion problems involving multiple time points, focus on what the question is actually asking. Average velocity measures the overall displacement divided by total time, regardless of what happens in between. To find average velocity, you need the displacement (change in position) and total time interval. The particle starts at

x = 5

m at

t = 0

s and ends at

x = 9

m at

t = 4

s. The displacement is

\Delta x = 9 - 5 = 4

m, and the time interval is

\Delta t = 4 - 0 = 4

s. Therefore, average velocity =

\frac{4 \text{ m}}{4 \text{ s}} = 1

m/s. Choice A (1 m/s) is correct based on this calculation. Choice B (2 m/s) might result from incorrectly using the maximum displacement between any two points (like from

t = 2

s to

t = 4

s, where displacement is 8 m over 2 s). Choice C (-1 m/s) could come from mistakenly calculating the velocity for just the first segment (from

t = 0

t = 2

s, where displacement is -4 m over 2 s). Choice D (0 m/s) might occur if you incorrectly think that because the particle changes direction, the average velocity is zero. The key insight is that average velocity only cares about the starting and ending positions over the specified time interval. The position at

t = 2

s is completely irrelevant to this calculation—it's included to test whether you understand that average velocity ignores intermediate motion details.

Question 15

A particle moves along the x-axis. Its position is given by $x(t) = 2t^3 - 12t^2 + 18t + 5$ , where $x$ is in meters and $t$ is in seconds. What is the acceleration of the particle at $t = 2$ seconds?

$0$ m/s² (correct answer)
$12$ m/s²
$-24$ m/s²
$-12$ m/s²
$-6$ m/s²

Explanation: When you encounter a kinematics problem with a position function, remember that acceleration is the second derivative of position with respect to time. This connects the mathematical concept of derivatives to the physical motion of objects. To find acceleration, you need to differentiate the position function twice. Starting with

x(t) = 2t^3 - 12t^2 + 18t + 5

, the first derivative gives velocity:

v(t) = \frac{dx}{dt} = 6t^2 - 24t + 18

. The second derivative gives acceleration:

a(t) = \frac{dv}{dt} = 12t - 24

. At

t = 2

seconds:

a(2) = 12(2) - 24 = 24 - 24 = 0

m/s². Looking at the wrong answers: Choice B (

12

m/s²) represents a common error where students substitute

t = 2

into the coefficient

12t

without completing the subtraction. Choice C (

-24

m/s²) occurs when students use only the constant term from the acceleration function, ignoring the

12t

component entirely. Choice D (

-12

m/s²) might result from sign errors or incorrectly applying the derivatives. The correct answer is A:

0

m/s². Study tip: Always remember the derivative chain for kinematics: position → velocity → acceleration. When finding acceleration at a specific time, substitute that time value into your second derivative function, not into intermediate steps. Practice taking derivatives of polynomial functions until this becomes automatic, as kinematics problems frequently test your calculus skills in physics contexts.

Question 16

An object moves with constant acceleration. Its velocity changes from $8$ m/s to $20$ m/s over a distance of $14$ m. What is the acceleration of the object?

$12$ m/s² (correct answer)
$6$ m/s²
$3$ m/s²
$1.5$ m/s²
$24$ m/s²

Explanation: When you encounter kinematics problems involving constant acceleration, you have three key equations at your disposal. The challenge is selecting the right one based on what information you're given and what you need to find. Here, you know initial velocity (

v_i = 8

m/s), final velocity (

v_f = 20

m/s), and distance (

d = 14

m), and you need acceleration. The perfect equation is

v_f^2 = v_i^2 + 2ad

. Substituting your values:

(20)^2 = (8)^2 + 2a(14)

This gives you:

400 = 64 + 28a

Solving for acceleration:

336 = 28a

, so

a = 12

m/s² This confirms answer A is correct. Let's examine why the other options are wrong. Option B (

6

m/s²) would result if you mistakenly used

a = \frac{v_f - v_i}{t}

but incorrectly calculated time as 2 seconds (which isn't given). Option C (

3

m/s²) might come from incorrectly using

a = \frac{v_f - v_i}{2d}

, treating distance like time. Option D (

1.5

m/s²) could result from various algebraic errors, such as forgetting to square the velocities or miscalculating the arithmetic. Study tip: Always identify what variables you know and what you need to find, then select the kinematic equation that connects them directly. This prevents you from making unnecessary assumptions about unknown quantities like time, which isn't provided in this problem.

Question 17

A particle moves along a circular path with constant speed $v$ . Which statement correctly describes the particle's acceleration?

The acceleration is constant in magnitude and directed toward the center of the circle (correct answer)
The acceleration is zero because the speed is constant
The acceleration is constant in both magnitude and direction
The acceleration varies in magnitude but is always tangent to the circle
The acceleration is zero except when the particle changes direction

Explanation: When you encounter circular motion problems, remember that acceleration depends on changes in velocity, and velocity is a vector quantity that includes both magnitude and direction. For a particle moving in a circle at constant speed, while the magnitude of velocity stays the same, the direction continuously changes. This changing direction means the velocity vector is changing, which requires acceleration. This acceleration is called centripetal acceleration, with magnitude

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

and direction always pointing toward the center of the circle. Choice A is correct because the acceleration has constant magnitude (since speed

v

and radius

r

are constant) and always points toward the center. Though the acceleration vector itself rotates as the particle moves around the circle, its magnitude and inward direction remain consistent. Choice B reflects a common misconception that constant speed means zero acceleration. This ignores that acceleration results from any change in the velocity vector, including direction changes. Choice C is incorrect because while the acceleration's magnitude stays constant, its direction continuously changes as it always points toward the center from the particle's current position. Choice D incorrectly describes tangential acceleration, which occurs when speed changes. Since the speed is constant here, there's no tangential component—only the centripetal component pointing inward. Remember this key distinction: constant speed in circular motion means zero tangential acceleration but non-zero centripetal acceleration. Always consider both the magnitude and direction of vectors when analyzing motion problems.

Question 18

Based on the velocity-time graph shown, what is the displacement of the object between $t = 2$ s and $t = 6$ s?

$16$ m (correct answer)
$8$ m
$12$ m
$20$ m
$4$ m

Explanation: Displacement equals the area under the velocity-time curve. From

t = 2

s to

t = 6

s (4-second interval), the velocity is constant at

4

m/s. Area = base × height =

4 \text{ s} \times 4 \text{ m/s} = 16

m. Choice B uses half the time interval. Choice C uses average of initial and final velocities incorrectly. Choice D adds the velocity value to the displacement. Choice E uses only the velocity value without considering time.

Question 19

A ball is dropped from rest from a height $h$ above the ground. At the same instant, another identical ball is thrown horizontally from the same height with initial speed $v_0$ . Which statement correctly describes their motion?

Both balls have the same vertical acceleration, but the thrown ball reaches the ground first due to its horizontal motion
Both balls have the same vertical acceleration and reach the ground simultaneously, but with different speeds (correct answer)
The dropped ball has greater vertical acceleration and reaches the ground first
Both balls reach the ground simultaneously with identical velocities

Explanation: Both balls experience only gravitational acceleration in the vertical direction (

a_y = -g

), regardless of horizontal motion. Since both start from the same height with zero initial vertical velocity, they fall for the same time

t = \sqrt{\frac{2h}{g}}

. However, when they reach the ground, the dropped ball has only vertical velocity

v_y = -\sqrt{2gh}

, while the thrown ball has both horizontal velocity

v_x = v_0

and the same vertical velocity, giving it a larger total speed

v = \sqrt{v_0^2 + 2gh}

. Choice A incorrectly suggests horizontal motion affects fall time. Choice C incorrectly suggests different vertical accelerations. Choice D incorrectly suggests identical final velocities.