What is Kinematics on the Physics Regents?

Kinematics is a Mechanics topic on the Physics Regents. RegentsLab has 68 practice questions on it, each with a full worked solution.

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Kinematics: Physics Regents Practice

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Question 4 · 1 pt

A car travels $150\ \mathrm{m}$ in $10\ \mathrm{s}$ at constant speed. What is its speed?

(A) $160\ \mathrm{m/s}$
(B) $15\ \mathrm{m/s}$ ✓
(C) $0.067\ \mathrm{m/s}$
(D) $1500\ \mathrm{m/s}$

Worked solution

$v = \frac{d}{t} = \frac{150}{10}$ Speed = distance / time.

$= 15\ \mathrm{m/s}$

Answer: 15\ \mathrm{m/s}

Average speed is the distance traveled divided by the time taken.

Question 1 · 2 pts

A car travels a distance of $150\ \text{m}$ in a time of $5.0\ \text{s}$ while moving at constant speed. What is the average speed of the car?

(A) $145\ \text{m/s}$
(B) $30\ \text{m/s}$ ✓
(C) $750\ \text{m/s}$
(D) $0.033\ \text{m/s}$

Worked solution

$\bar{v} = \frac{d}{t}$ Average speed relationship from the reference tables.

$\bar{v} = \frac{150\,\text{m}}{5.0\,\text{s}}$

$\bar{v} = 30\,\text{m/s}$

Answer: $30\ \text{m/s}$

Average speed equals total distance divided by total elapsed time, $\bar{v}=d/t$.

Question 2 · 2 pts

A car speeds up uniformly from $10\ \text{m/s}$ to $30\ \text{m/s}$ in a time of $4.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) $10\ \text{m/s}^2$
(B) $80\ \text{m/s}^2$
(C) $7.5\ \text{m/s}^2$
(D) $5.0\ \text{m/s}^2$ ✓

Worked solution

$a = \frac{\Delta v}{t} = \frac{v_f - v_i}{t}$

$a = \frac{30\,\text{m/s} - 10\,\text{m/s}}{4.0\,\text{s}} = \frac{20\,\text{m/s}}{4.0\,\text{s}}$

$a = 5.0\,\text{m/s}^2$

Answer: $5.0\ \text{m/s}^2$

Acceleration is the rate of change of velocity, $a=\Delta v/t$, so only the change in velocity divided by time matters.

Question 10 · 2 pts

A position-time graph for an object moving in a straight line is a straight line passing through the points $(0\text{ s}, 4\text{ m})$ and $(5\text{ s}, 24\text{ m})$. What is the object's velocity?

(A) 4 m/s ✓
(B) 5.6 m/s
(C) 0.25 m/s
(D) 4.8 m/s

Worked solution

$v = \frac{\Delta d}{\Delta t} = \frac{d_f - d_i}{t_f - t_i}$ Velocity is the slope of a position-time graph.

$v = \frac{24 - 4}{5 - 0} = \frac{20}{5} = 4\ \text{m/s}$

Answer: 4 m/s

On a position-time graph the velocity is the slope, $v=\frac{\Delta d}{\Delta t}$, so the starting position must be subtracted from the final position.

Question 16 · 2 pts

An object starts from rest. Its acceleration-time graph shows a constant acceleration of $4\text{ m/s}^2$ for the first $3\text{ s}$, followed by a constant acceleration of $2\text{ m/s}^2$ for the next $4\text{ s}$. What is the object's speed at the end of the $7\text{ s}$?

(A) 12 m/s
(B) 14 m/s
(C) 20 m/s ✓
(D) 21 m/s

Worked solution

$\Delta v = \text{area under the } a\text{-}t \text{ graph}$ Change in velocity equals the area under the acceleration-time curve.

$\Delta v_1 = (4)(3) = 12\ \text{m/s},\quad \Delta v_2 = (2)(4) = 8\ \text{m/s}$

$v = 0 + 12 + 8 = 20\ \text{m/s}$

Answer: 20 m/s

On an acceleration-time graph the area under the curve equals the change in velocity, so the areas of both segments are added to the initial speed.

Question 1 · 2 pts

A car starts from rest and accelerates uniformly, reaching a speed of $12\ \text{m/s}$ in $4.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) $48\ \text{m/s}^2$
(B) $8.0\ \text{m/s}^2$
(C) $0.33\ \text{m/s}^2$
(D) $3.0\ \text{m/s}^2$ ✓

Worked solution

$a = \dfrac{v_f - v_i}{t}$ Definition of average acceleration

$a = \dfrac{12\ \text{m/s} - 0}{4.0\ \text{s}}$

$a = 3.0\ \text{m/s}^2$

Answer: $3.0\ \text{m/s}^2$

Acceleration is the change in velocity divided by the time interval; dividing the 12 m/s change in speed by the 4.0 s interval gives 3.0 m/s^2.

Question 13 · 2 pts

A car starting from rest accelerates uniformly and reaches a speed of $20\text{ m/s}$ after $4.0\text{ s}$. What is the magnitude of the car's acceleration?

(A) $0.20\text{ m/s}^2$
(B) $16\text{ m/s}^2$
(C) $80\text{ m/s}^2$
(D) $5.0\text{ m/s}^2$ ✓

Worked solution

$a = \dfrac{v_f - v_i}{t}$ Definition of average acceleration

$a = \dfrac{20\text{ m/s} - 0}{4.0\text{ s}}$

$a = 5.0\text{ m/s}^2$

Answer: $5.0\text{ m/s}^2$

Acceleration is the change in velocity divided by the time interval. Since the car starts from rest, the change in velocity equals the final velocity, so dividing 20 m/s by 4.0 s gives 5.0 m/s^2.

Question 26 · 2 pts

A car starting from rest accelerates uniformly and reaches a speed of $24\ \text{m/s}$ in $6.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) 4.0 m/s² ✓
(B) 144 m/s²
(C) 24 m/s²
(D) 0.25 m/s²

Worked solution

$a = \dfrac{\Delta v}{\Delta t}$ Definition of average acceleration

$a = \dfrac{24\ \text{m/s} - 0}{6.0\ \text{s}}$

$a = 4.0\ \text{m/s}^2$

Answer: 4.0 m/s²

Acceleration is the change in velocity divided by the time interval; since the car starts from rest, the change in velocity equals the final speed of 24 m/s, giving a = 24/6.0 = 4.0 m/s².

Question 39 · 2 pts

A stone is dropped from rest and falls freely under gravity for 3.0 seconds before striking the ground. What is the speed of the stone just before it strikes the ground? (Neglect air resistance; g = 9.81 m/s^2.)

(A) 20 m/s
(B) 29 m/s ✓
(C) 9.8 m/s
(D) 44 m/s

Worked solution

$v = v_0 + g t$ free fall from rest, v0 = 0

$v = (9.81\ \text{m/s}^2)(3.0\ \text{s})$

$v = 29.43\ \text{m/s} \approx 29\ \text{m/s}$

Answer: 29 m/s

For an object dropped from rest, final speed equals g times the fall time; multiplying 9.81 m/s^2 by 3.0 s gives about 29 m/s. Choosing v = g (9.8 m/s) ignores the time factor, a common error.

Question 51 · 4 pts · constructed response

A car starting from rest accelerates uniformly and reaches a speed of $24\ \text{m/s}$ in $6.0\ \text{s}$. Calculate the car's acceleration and the distance it travels during this time.

Model answer: a = 4.0 m/s^2; d = 72 m

Worked solution

$a = \dfrac{v_f - v_0}{t} = \dfrac{24 - 0}{6.0}$ acceleration from change in velocity

$a = 4.0\ \text{m/s}^2$

$d = v_0 t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4.0)(6.0)^2$ using kinematics equation

$d = 72\ \text{m}$

Answer: a = 4.0 m/s^2; d = 72 m

Since the car starts from rest, acceleration is found directly from the change in velocity over time, and the distance follows from the kinematics equation with initial velocity zero.

Question 59 · 4 pts · constructed response

A ball is thrown horizontally from the top of a cliff $19.6\ \text{m}$ high with an initial horizontal speed of $15\ \text{m/s}$. Calculate the time it takes the ball to reach the ground and the horizontal distance it travels before landing.

Model answer: The ball falls for $2.0\ \text{s}$ and lands $30\ \text{m}$ from the base of the cliff.

Worked solution

$h = \tfrac{1}{2}gt^2$ vertical motion, initial vertical velocity is zero

$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2(19.6\ \text{m})}{9.81\ \text{m/s}^2}} = \sqrt{3.996\ \text{s}^2} \approx 2.0\ \text{s}$

$x = v_x t = (15\ \text{m/s})(2.0\ \text{s}) = 30\ \text{m}$

Answer: The ball falls for $2.0\ \text{s}$ and lands $30\ \text{m}$ from the base of the cliff.

In projectile motion the horizontal and vertical motions are independent; the fall time depends only on the drop height and $g$, and the horizontal range is simply the constant horizontal speed multiplied by that fall time.

Question 1 · 2 pts

A car starts from rest and speeds up uniformly to a velocity of 20. m/s in 4.0 s. What is the magnitude of the car's acceleration?

(A) 0.20 m/s^2
(B) 5.0 m/s^2 ✓
(C) 80. m/s^2
(D) 24 m/s^2

Worked solution

$a = \dfrac{\Delta v}{\Delta t}$ Definition of average acceleration

$a = \dfrac{20.\text{ m/s} - 0}{4.0\text{ s}}$

$a = 5.0\text{ m/s}^2$

Answer: 5.0 m/s^2

Acceleration is the change in velocity divided by the time interval over which it occurs. Dividing the 20. m/s change in speed by the 4.0 s time interval gives 5.0 m/s^2; the distractors come from inverting the ratio or from combining the numbers with the wrong operation.

Question 13 · 2 pts

A car starts from rest and accelerates uniformly at $2.5\ \text{m/s}^2$ for $6.0\ \text{s}$. What is the distance the car travels during this time?

(A) 15 m
(B) 7.5 m
(C) 45 m ✓
(D) 90 m

Worked solution

$d = v_0 t + \tfrac{1}{2}at^2$ starts from rest so $v_0 = 0$

$d = \tfrac{1}{2}(2.5\ \text{m/s}^2)(6.0\ \text{s})^2$

$d = \tfrac{1}{2}(2.5)(36) = 45\ \text{m}$

Answer: 45 m

Since the car starts from rest, the displacement is found using $d=\tfrac12at^2$. Forgetting the factor of one-half doubles the answer to 90 m, a common error.

Question 25 · 2 pts

A car starts from rest and accelerates uniformly, reaching a speed of $20\ \text{m/s}$ in $5.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) $5.0\ \text{m/s}^2$
(B) $4.0\ \text{m/s}^2$ ✓
(C) $100\ \text{m/s}^2$
(D) $0.25\ \text{m/s}^2$

Worked solution

$a = \dfrac{\Delta v}{\Delta t}$ Definition of average acceleration

$a = \dfrac{20\ \text{m/s} - 0}{5.0\ \text{s}}$

$a = 4.0\ \text{m/s}^2$

Answer: $4.0\ \text{m/s}^2$

Acceleration is the change in velocity divided by the time interval over which the change occurs; since the car starts from rest, $\Delta v$ equals the final speed.

Question 37 · 2 pts

A car starting from rest accelerates uniformly at 3.0 m/s^2 for 4.0 s. What total distance does the car travel during this time?

(A) 48 m
(B) 12 m
(C) 6.0 m
(D) 24 m ✓

Worked solution

$d = v_i t + \tfrac{1}{2}at^2$ Kinematics equation for constant acceleration from rest

$d = (0)(4.0) + \tfrac{1}{2}(3.0)(4.0)^2$

$d = \tfrac{1}{2}(3.0)(16) = 24\ \text{m}$

Answer: 24 m

Starting from rest, the distance under uniform acceleration is one-half times a times t squared; with a = 3.0 m/s^2 and t = 4.0 s, d = 0.5 x 3.0 x 16 = 24 m.

Question 51 · 4 pts · constructed response

A cart starts from rest and accelerates uniformly at $3.0\ \text{m/s}^2$ for $5.0\ \text{s}$. Calculate the total distance the cart travels during this $5.0\ \text{s}$ interval.

Model answer: The cart travels 37.5 m, which rounds to 38 m.

Worked solution

$d = v_0 t + \tfrac{1}{2}at^2$ Kinematics equation for displacement with constant acceleration

$d = (0)(5.0) + \tfrac{1}{2}(3.0)(5.0)^2$

$d = \tfrac{1}{2}(3.0)(25)$

$d = 37.5\ \text{m} \approx 38\ \text{m}$

Answer: The cart travels 37.5 m, which rounds to 38 m.

Since the cart starts from rest, the initial velocity term drops out and displacement is found entirely from the acceleration term of the kinematics equation.

Question 59 · 4 pts · constructed response

A car starts from rest and accelerates uniformly, reaching a speed of $24\ \text{m/s}$ in $6.0\ \text{s}$. Calculate the car's acceleration and the distance it travels during this time.

Model answer: a = 4.0 m/s^2 and the car travels 72 m.

Worked solution

$a = \dfrac{v_f - v_i}{t} = \dfrac{24 - 0}{6.0}$ Use the definition of acceleration

$a = 4.0\ \text{m/s}^2$

$d = v_i t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4.0)(6.0)^2$ Use kinematics equation for displacement

$d = 72\ \text{m}$

Answer: a = 4.0 m/s^2 and the car travels 72 m.

With constant acceleration from rest, the acceleration is found directly from the change in velocity over time, and the distance follows from the standard kinematics displacement equation using that acceleration and time.

Question 67 · 4 pts · constructed response

A car traveling at 30. m/s applies its brakes and decelerates uniformly to rest in 5.0 s. Calculate the magnitude of the car's acceleration and the distance it travels while stopping.

Model answer: The magnitude of the acceleration is 6.0 m/s^2 and the car travels 75 m.

Worked solution

$a = \dfrac{v_f - v_i}{t} = \dfrac{0 - 30.}{5.0} = -6.0\ \text{m/s}^2$ Definition of acceleration

$|a| = 6.0\ \text{m/s}^2$

$d = \dfrac{v_i + v_f}{2}\,t = \dfrac{30. + 0}{2}(5.0) = 75\ \text{m}$ Distance from average velocity

Answer: The magnitude of the acceleration is 6.0 m/s^2 and the car travels 75 m.

The acceleration is the change in velocity over time, (0 - 30.)/5.0 = -6.0 m/s^2, so its magnitude is 6.0 m/s^2; the stopping distance is the average velocity times the time, (30./2)(5.0) = 75 m.

Question 15 · 2 pts

A car starting from rest accelerates uniformly and reaches a speed of $20.\ \text{m/s}$ in $5.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) $4.0\ \text{m/s}^2$ ✓
(B) $20.\ \text{m/s}^2$
(C) $100\ \text{m/s}^2$
(D) $0.25\ \text{m/s}^2$

Worked solution

$a = \dfrac{\Delta v}{\Delta t}$ definition of average acceleration

$a = \dfrac{20.\ \text{m/s} - 0\ \text{m/s}}{5.0\ \text{s}}$

$a = 4.0\ \text{m/s}^2$

Answer: $4.0\ \text{m/s}^2$

Acceleration is the change in velocity divided by the time interval over which the change occurs; here the car's velocity changes by 20. m/s in 5.0 s, giving 4.0 m/s^2.

Question 28 · 2 pts

A car starts from rest and accelerates uniformly, reaching a speed of $24\ \text{m/s}$ in $6.0\ \text{s}$. What is the magnitude of the car's acceleration?

(A) 4.0 m/s^2 ✓
(B) 24 m/s^2
(C) 0.25 m/s^2
(D) 144 m/s^2

Worked solution

$a = \dfrac{\Delta v}{\Delta t}$ definition of average acceleration

$a = \dfrac{24\ \text{m/s} - 0}{6.0\ \text{s}}$

$a = 4.0\ \text{m/s}^2$

Answer: 4.0 m/s^2

Acceleration is the change in velocity divided by the time interval; dividing 24 m/s by 6.0 s gives 4.0 m/s^2. Students who invert the ratio get 0.25 m/s^2, and those who multiply instead of divide get 144 m/s^2.

Question 41 · 2 pts

A car starting from rest accelerates uniformly and reaches a speed of $24\ \text{m/s}$ in $6.0\ \text{s}$. What distance does the car travel during this time?

(A) 96 m
(B) 36 m
(C) 72 m ✓
(D) 144 m

Worked solution

$a = \dfrac{\Delta v}{t} = \dfrac{24\ \text{m/s} - 0}{6.0\ \text{s}} = 4.0\ \text{m/s}^2$ Find acceleration

$d = \bar{v}t = \left(\dfrac{v_0+v_f}{2}\right)t = \left(\dfrac{0+24}{2}\right)(6.0)$ Use average velocity

$d = (12\ \text{m/s})(6.0\ \text{s}) = 72\ \text{m}$

Answer: 72 m

For uniform acceleration from rest, the distance traveled equals the average velocity multiplied by the time. The average of 0 and 24 m/s is 12 m/s, giving a distance of 72 m. Choosing 144 m is the common error of using the final velocity instead of the average velocity in $d = vt$.

Question 51 · 4 pts · constructed response

A car starting from rest accelerates uniformly and reaches a speed of $24.0\ \text{m/s}$ in $6.00\ \text{s}$. Calculate the car's acceleration and the total distance it travels during this $6.00\ \text{s}$ interval.

Model answer: The acceleration is $4.00\ \text{m/s}^2$ and the distance traveled is $72.0\ \text{m}$.

Worked solution

$a = \dfrac{v_f - v_i}{t} = \dfrac{24.0 - 0}{6.00}$ Use the definition of acceleration

$a = 4.00\ \text{m/s}^2$

$d = v_i t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4.00)(6.00)^2$ Use kinematics with v_i = 0

$d = 72.0\ \text{m}$

Answer: The acceleration is $4.00\ \text{m/s}^2$ and the distance traveled is $72.0\ \text{m}$.

Since the car starts from rest and accelerates uniformly, the acceleration is found directly from the change in velocity over time, and that acceleration is then substituted into the kinematics displacement equation to find the distance traveled.

Question 59 · 4 pts · constructed response

A car starting from rest accelerates uniformly and reaches a speed of 24 m/s in 6.0 s. Calculate the magnitude of the car's acceleration, and calculate the distance the car travels during this 6.0 s interval.

Model answer: a = 4.0 m/s^2; d = 72 m

Worked solution

$a = \dfrac{v_f - v_i}{t} = \dfrac{24\ \text{m/s} - 0}{6.0\ \text{s}}$ Use the definition of average acceleration

$a = 4.0\ \text{m/s}^2$

$d = v_i t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(4.0\ \text{m/s}^2)(6.0\ \text{s})^2$ Use kinematics equation for displacement

$d = \tfrac{1}{2}(4.0)(36) = 72\ \text{m}$

Answer: a = 4.0 m/s^2; d = 72 m

Since the car accelerates uniformly from rest, its acceleration is simply the change in velocity divided by the time interval, and the distance traveled follows directly from the kinematics equation relating displacement to initial velocity, acceleration, and time.

Question 67 · 4 pts · constructed response

A ball is thrown horizontally with an initial speed of $12\ \text{m/s}$ from the top of a cliff $19.62\ \text{m}$ high. Calculate the time it takes the ball to reach the ground, and calculate the horizontal distance the ball travels during its fall. (Neglect air resistance.)

Model answer: The ball falls for $2.0\ \text{s}$ and travels a horizontal distance of $24\ \text{m}$.

Worked solution

$h = \frac{1}{2}gt^2$ Vertical motion is independent of horizontal motion

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(19.62\ \text{m})}{9.81\ \text{m/s}^2}} = \sqrt{4.0\ \text{s}^2} = 2.0\ \text{s}$

$x = v_0 t = (12\ \text{m/s})(2.0\ \text{s}) = 24\ \text{m}$ Horizontal velocity is constant since there is no horizontal acceleration

Answer: The ball falls for $2.0\ \text{s}$ and travels a horizontal distance of $24\ \text{m}$.

In projectile motion the vertical and horizontal components of motion are independent. The time to fall is found solely from the vertical drop using kinematics with $g$, and that time is then used with the constant horizontal velocity to find the range.

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