Differential Equations Quiz: Falling Objects With Drag

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Falling Objects With DragQuestion 1 of 19

An object of mass $m$ and drag coefficient $k$ is dropped from rest in a fluid. Let $v_T = mg/k$ be the terminal velocity and $\tau = m/k$ be the time constant of the system. What is the total distance the object falls in a time interval equal to two time constants, i.e., $t = 2\tau$ ?

v_T \tau (1 + e^{-2})

v_T \tau (2 + e^{-2})

v_T (1 - e^{-2})

2 v_T \tau

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Differential Equations Quiz

Differential Equations Quiz: Falling Objects With Drag

Practice Falling Objects With Drag in Differential Equations 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 Falling Objects With Drag, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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

$v_T \tau (1 + e^{-2})$ (correct answer)
$v_T \tau (2 + e^{-2})$
$v_T (1 - e^{-2})$
$2 v_T \tau$

Explanation: First, find the velocity function:

v(t) = v_T(1 - e^{-t/\tau})

. To find the distance fallen, integrate the velocity from

t=0

t=2\tau

. Let

x(t)

be the distance.

x(t) = \int_0^t v(s) ds = \int_0^t v_T(1 - e^{-s/\tau}) ds = v_T [s + \tau e^{-s/\tau}]_0^t = v_T((t + \tau e^{-t/\tau}) - (0 + \tau e^0)) = v_T(t + \tau e^{-t/\tau} - \tau)

. Now, substitute

t = 2\tau

x(2\tau) = v_T(2\tau + \tau e^{-2\tau/\tau} - \tau) = v_T(\tau + \tau e^{-2}) = v_T \tau (1 + e^{-2})

. Distractor B results from an integration constant error. Distractor C represents the velocity at

t=2\tau

, not the distance. Distractor D is the distance that would be covered if the object traveled at its terminal velocity for the entire duration.

Question 2

The velocity of a falling object with linear drag is $v(t) = v_T(1 - e^{-t/\tau})$ , where $v_T$ is terminal velocity and $\tau=m/k$ is the time constant. Which of the following statements provides the best physical interpretation of the time constant $\tau$ ?

$\tau$ is the time required for the object to reach approximately 99% of its terminal velocity.
$\tau$ is the time at which the object reaches half of its terminal velocity.
If the object maintained its initial acceleration, it would reach its terminal velocity $v_T$ in time $\tau$ . (correct answer)
$\tau$ is the time at which the acceleration of the object is half of its initial value.

Explanation: The governing differential equation is

dv/dt = g - (k/m)v

. The initial acceleration (at

t=0

, when

v=0

) is

a(0) = g

. If the object were to continue accelerating at this constant rate, its velocity would be

v(t) = gt

. To reach terminal velocity,

v_T = mg/k

, would require time

t

such that

mg/k = gt

. Solving for

t

gives

t = m/k

, which is defined as the time constant

\tau

. So,

\tau

is the time it would take to reach

v_T

with constant initial acceleration

g

. Choice A is incorrect; that would be roughly

t=5\tau

. Choice B is incorrect; half of terminal velocity is reached at

t = \tau \ln(2)

. Choice D is incorrect; acceleration is

a(t)=ge^{-t/\tau}

, so it is halved when

e^{-t/\tau}=0.5

, which is at

t=\tau \ln(2)

Question 3

A particle of mass $m$ is projected downward into a fluid with an initial speed of $10$ m/s. The net downward force on the particle is given by $F_{net} = mg - B - kv$ , where $B$ is a constant buoyant force and $k$ is a positive drag coefficient. The particle's terminal velocity is observed to be $v_T = 5$ m/s. Which of the following describes the subsequent motion of the particle?

The particle will speed up, with its velocity approaching infinity.
The particle will maintain a constant speed of $10$ m/s as there is no net force.
The particle will slow down, with its speed approaching $5$ m/s from above. (correct answer)
The particle will slow down to a complete stop, then reverse direction and move upward.

Explanation: The equation of motion is

m(dv/dt) = (mg-B) - kv

. The terminal velocity

v_T

is the speed at which the net force is zero, so

mg - B - kv_T = 0

. This implies that the term

(mg-B)

is equal to

kv_T

. Substituting this back into the equation of motion gives

m(dv/dt) = kv_T - kv = k(v_T - v)

. The initial velocity is

v(0) = 10

m/s, and the terminal velocity is

v_T = 5

m/s. Since

v(0) > v_T

, the term

(v_T - v)

is initially negative. This means

dv/dt

is negative, so the particle decelerates. The velocity will decrease from

10

m/s, asymptotically approaching the stable equilibrium velocity

v_T = 5

m/s.

Question 4

The velocity of an object falling from rest with linear air resistance is given by the function $v(t) = A(1 - e^{-Bt})$ , where $A$ and $B$ are positive constants. Which of the following correctly identifies the physical meaning of these constants in terms of the object's mass $m$ , the acceleration due to gravity $g$ , and the drag coefficient $k$ ?

$A$ is the initial acceleration $g$ , and $B$ is the time constant $m/k$ .
$A$ is the terminal velocity $mg/k$ , and $B$ is the reciprocal of the time constant, $k/m$ . (correct answer)
$A$ is the terminal velocity $mg/k$ , and $B$ is the time constant $m/k$ .
$A$ is the mass $m$ , and $B$ is the drag coefficient $k$ .

Explanation: The standard solution to the differential equation

m(dv/dt) = mg - kv

with initial condition

v(0)=0

v(t) = \frac{mg}{k}(1 - e^{-(k/m)t})

. Comparing this to the given form

v(t) = A(1 - e^{-Bt})

, we can identify the constants by matching the terms. The pre-factor

A

corresponds to

\frac{mg}{k}

, which is the terminal velocity

v_T

. The constant

B

in the exponent corresponds to

\frac{k}{m}

. The time constant is defined as

\tau = m/k

, so

B = 1/\tau

. Therefore,

A

is the terminal velocity and

B

is the ratio of the drag coefficient to the mass (the reciprocal of the time constant).

Question 5

An object of mass $m=5$ kg is falling with a speed of $v_0 = 10$ m/s at time $t=0$ . It is subject to gravity ( $g=10$ m/s²) and a linear drag force $F_d = -2.5v$ . What is the speed of the object at $t=2$ seconds?

$20 - 20e^{-1}$ m/s
$20 + 10e^{-1}$ m/s
$20 - 10e^{-4}$ m/s
$20 - 10e^{-1}$ m/s (correct answer)

Explanation: The equation of motion is

m(dv/dt) = mg - kv

. Plugging in the values, we have

5(dv/dt) = 5(10) - 2.5v

, which simplifies to

dv/dt = 10 - 0.5v

. The terminal velocity is

v_T = mg/k = 50/2.5 = 20

m/s. The general solution is of the form

v(t) = v_T + C e^{-(k/m)t}

. Here,

k/m = 2.5/5 = 0.5

, so

v(t) = 20 + C e^{-0.5t}

. We use the initial condition

v(0)=10

to find

C

10 = 20 + C e^0

, which gives

C = -10

. So, the specific solution is

v(t) = 20 - 10e^{-0.5t}

. To find the speed at

t=2

, we evaluate

v(2) = 20 - 10e^{-0.5(2)} = 20 - 10e^{-1}

. Choice A comes from incorrectly using

v(0)=0

. Choice C comes from incorrectly using

m/k

instead of

k/m

in the exponent.

Question 6

A small sphere is dropped from rest in a viscous fluid. It is observed to reach 50% of its terminal velocity in $\ln(2)$ seconds. Assuming the drag force is linear with velocity ( $F_d = -kv$ ), what is the time constant $\tau = m/k$ for this system?

$1$ s (correct answer)
$\ln(2)$ s
$2$ s
$1/\ln(2)$ s

Explanation: The velocity of an object falling from rest with linear drag is given by the equation

v(t) = v_T(1 - e^{-t/\tau})

, where

v_T

is the terminal velocity and

\tau

is the time constant. We are given that at

t = \ln(2)

, the velocity is

v(t) = 0.5 v_T

. Substituting these values into the equation gives:

0.5 v_T = v_T(1 - e^{-\ln(2)/\tau})

. Dividing by

v_T

, we get

0.5 = 1 - e^{-\ln(2)/\tau}

. Rearranging yields

e^{-\ln(2)/\tau} = 0.5

. Taking the natural logarithm of both sides, we have

-\ln(2)/\tau = \ln(0.5) = -\ln(2)

. Therefore,

\ln(2)/\tau = \ln(2)

, which implies that

\tau = 1

second.

Question 7

Two spherical objects, Sphere 1 and Sphere 2, are made of the same uniform material. Sphere 1 has twice the radius of Sphere 2 ( $r_1 = 2r_2$ ). They are dropped from rest in the same fluid, where the drag force is linear with velocity, $F_d = -kv$ , and the drag coefficient $k$ is directly proportional to the radius $r$ (i.e., $k=cr$ for some constant $c$ ). How do their terminal velocities, $v_{T1}$ and $v_{T2}$ , compare?

$v_{T1} = 2v_{T2}$
$v_{T1} = 4v_{T2}$ (correct answer)
$v_{T1} = 8v_{T2}$
$v_{T1} = \sqrt{2}v_{T2}$

Explanation: The terminal velocity is given by the formula

v_T = mg/k

. The mass

m

of a sphere is its density

\rho

times its volume

V = \frac{4}{3}\pi r^3

. Since the material is the same,

\rho

is constant, so

m

is proportional to

r^3

. Thus,

m_1/m_2 = (r_1/r_2)^3 = 2^3 = 8

. We are given that the drag coefficient

k

is proportional to the radius

r

. Thus,

k_1/k_2 = r_1/r_2 = 2

. Now we can find the ratio of the terminal velocities:

\frac{v_{T1}}{v_{T2}} = \frac{m_1 g / k_1}{m_2 g / k_2} = \frac{m_1}{m_2} \cdot \frac{k_2}{k_1}

. Substituting the ratios we found:

\frac{v_{T1}}{v_{T2}} = (8) \cdot \frac{1}{2} = 4

. Therefore,

v_{T1} = 4v_{T2}

Question 8

An object is dropped from rest, and its velocity is described by $v(t) = v_T(1-e^{-t/\tau})$ , where $v_T$ is the terminal velocity and $\tau$ is the time constant. After a very long time ( $t \gg \tau$ ), the distance the object has fallen, $x(t)$ , can be approximated by a linear function $x(t) \approx At - B$ . What are the constants $A$ and $B$ ?

$A = v_T$ , $B = v_T \tau$ (correct answer)
$A = v_T$ , $B = 0$
$A = g$ , $B = v_T \tau$
$A = v_T$ , $B = \tau$

Explanation: To find the distance

x(t)

, we integrate the velocity function

v(t)

from

0

t

, with

x(0)=0

x(t) = \int_0^t v_T(1-e^{-s/\tau}) ds = v_T [s + \tau e^{-s/\tau}]_0^t = v_T ((t + \tau e^{-t/\tau}) - (0 + \tau e^0)) = v_T(t - \tau + \tau e^{-t/\tau})

. For very long times,

t \gg \tau

, the exponential term

e^{-t/\tau}

approaches zero. Therefore, the position function can be approximated as

x(t) \approx v_T(t - \tau) = v_T t - v_T \tau

. Comparing this to the form

At - B

, we find that

A = v_T

and

B = v_T \tau

. This approximation represents the distance the object would have traveled if it had been moving at terminal velocity for a time

t

, minus a 'head start' distance of

v_T \tau

Question 9

The velocity of an object of mass $m$ dropped from rest, subject to linear drag with coefficient $k$ , is given by $v(t) = \frac{mg}{k}(1 - e^{-kt/m})$ . What is the limiting value of this expression as the drag coefficient $k$ approaches zero (i.e., the case of no drag)?

$gt$ (correct answer)
$0$
$g$
The limit is infinite.

Explanation: To find the limit of

v(t)

k \to 0

, we can use L'Hôpital's Rule because the expression takes the indeterminate form

0/0

. Let's rewrite the expression as

v(t) = mg \cdot \frac{1 - e^{-kt/m}}{k}

. We evaluate the limit of the fraction:

\lim_{k\to 0} \frac{1 - e^{-kt/m}}{k}

. Differentiating the numerator and the denominator with respect to

k

, we get: Numerator derivative:

-(e^{-kt/m})(-t/m) = (t/m)e^{-kt/m}

. Denominator derivative:

1

. The limit becomes

\lim_{k\to 0} \frac{(t/m)e^{-kt/m}}{1} = (t/m)e^0 = t/m

. Therefore, the original limit is

mg \cdot (t/m) = gt

. This is the well-known formula for velocity under constant gravitational acceleration with no air resistance, which is the expected physical result.

Question 10

A skydiver of mass $m$ falls from rest. Air resistance is proportional to her velocity $v$ , with proportionality constant $k$ . At time $t_1$ , her speed is $v_1$ , and she opens her parachute, which instantly changes the drag coefficient to $k_2$ , where $k_2 > k$ . Which statement correctly describes her motion at the exact moment she opens the parachute?

Her velocity instantaneously drops to the new, lower terminal velocity $mg/k_2$ .
She immediately begins to decelerate, regardless of her speed $v_1$ when the parachute opened.
If her speed $v_1$ was greater than the new terminal velocity $mg/k_2$ , she will decelerate. (correct answer)
Her acceleration $dv/dt$ remains continuous at the moment the parachute opens.

Explanation: The differential equation for velocity is

m(dv/dt) = mg - kv

. At the moment the parachute opens, the drag coefficient

k

changes to

k_2

. The velocity

v_1

is continuous at this instant, but the acceleration changes to

dv/dt = g - (k_2/m)v_1

. The new terminal velocity is

v_{T2} = mg/k_2

. The skydiver will decelerate if

dv/dt < 0

, which means

g - (k_2/m)v_1 < 0

, or

mg < k_2v_1

. This is equivalent to

mg/k_2 < v_1

, or

v_{T2} < v_1

. Therefore, she decelerates only if her speed was greater than the new terminal velocity. Velocity cannot change instantaneously, so A is wrong. It's possible for

v_1

to be less than

v_{T2}

, in which case she would continue to accelerate (but more slowly), so B is wrong. Since

k

changes discontinuously, the acceleration

dv/dt = g - (k/m)v

must also be discontinuous, so D is wrong.

Question 11

A ball is dropped from rest in a medium where air resistance is proportional to velocity. The equation of motion is $\frac{dv}{dt} = g - kv$ , where $g = 10 \text{ m/s}^2$ and $k = 0.5 \text{ s}^{-1}$ . If the ball reaches 95% of its terminal velocity after time $t$ , what is the approximate value of $t$ ?

$t \approx 3.0 \text{ s}$
$t \approx 6.0 \text{ s}$ (correct answer)
$t \approx 9.0 \text{ s}$
$t \approx 12.0 \text{ s}$

Explanation: First find terminal velocity:

v_t = g/k = 10/0.5 = 20 \text{ m/s}

. The solution to the ODE is

v(t) = v_t(1 - e^{-kt}) = 20(1 - e^{-0.5t})

. For 95% of terminal velocity:

0.95 \times 20 = 20(1 - e^{-0.5t})

, so

0.95 = 1 - e^{-0.5t}

, giving

e^{-0.5t} = 0.05

. Taking natural log:

-0.5t = \ln(0.05) \approx -3.0

, so

t \approx 6.0 \text{ s}

. Choice A uses

\ln(0.05)

directly without dividing by

k

. Choice C incorrectly uses

\ln(0.95)

. Choice D doubles the correct answer due to sign error.

Question 12

Two identical spheres are dropped simultaneously from the same height, one in air ( $k_1 = 0.2 \text{ s}^{-1}$ ) and one in water ( $k_2 = 0.8 \text{ s}^{-1}$ ). Both experience linear drag proportional to velocity. After a long time, what is the ratio of their terminal velocities $v_{t,air}/v_{t,water}$ ?

$1:4$ (the sphere in air has 4 times the terminal velocity)
$4:1$ (the sphere in air has 4 times the terminal velocity) (correct answer)
$1:1$ (both spheres have the same terminal velocity)
$2:1$ (the sphere in air has 2 times the terminal velocity)

Explanation: Terminal velocity occurs when

mg = kv_t

, so

v_t = mg/k

. Since the spheres are identical, they have the same mass and experience the same gravitational force. Therefore

v_{t,air}/v_{t,water} = k_2/k_1 = 0.8/0.2 = 4

. The sphere in air has 4 times the terminal velocity because it experiences less drag resistance. Choice A inverts the ratio. Choice C ignores the different drag coefficients. Choice D uses the square root of the correct ratio.

Question 13

Consider the velocity-time relationship for a falling object with linear drag: $v(t) = v_t(1 - e^{-t/\tau})$ where $\tau = 1/k$ is the time constant. If the object reaches $63.2\%$ of its terminal velocity at time $t_1$ and $86.5\%$ at time $t_2$ , what is the relationship between $t_1$ and $t_2$ ?

$t_2 = 2t_1$ because the exponential relationship doubles the time interval (correct answer)
$t_2 = 1.5t_1$ because the velocity increase follows a power law
$t_2 = t_1 + \tau$ because each time constant adds the same increment
$t_2 = \sqrt{2}t_1$ because the approach to terminal velocity is quadratic

Explanation: At

63.2\% = 1 - 1/e \approx 0.632

0.632v_t = v_t(1 - e^{-t_1/\tau})

, so

e^{-t_1/\tau} = 0.368 = 1/e

, giving

t_1 = \tau

. At

86.5\% = 1 - 1/e^2 \approx 0.865

0.865v_t = v_t(1 - e^{-t_2/\tau})

, so

e^{-t_2/\tau} = 0.135 = 1/e^2

, giving

t_2 = 2\tau = 2t_1

. Choice B assumes linear relationship. Choice C incorrectly adds time constants. Choice D assumes quadratic approach.

Question 14

Two balls of different masses ( $m_1 = 2 \text{ kg}$ and $m_2 = 8 \text{ kg}$ ) but identical shape and size are dropped simultaneously from the same height in air. Both experience linear drag with the same coefficient $k = 0.5 \text{ s}^{-1}$ . After sufficient time, what is the ratio of their velocities $v_1/v_2$ ?

$v_1/v_2 = 1:1$ because terminal velocity is independent of mass
$v_1/v_2 = 4:1$ because lighter objects experience relatively more drag
$v_1/v_2 = 1:2$ because velocity scales with the square root of mass
$v_1/v_2 = 1:4$ because heavier objects fall faster in resistive media (correct answer)

Explanation: When you encounter problems involving objects falling through resistive media, the key insight is that terminal velocity depends on the balance between gravitational force and drag force. At terminal velocity, these forces are equal, so the object stops accelerating. For linear drag, the drag force is

F_d = kv

, where

k

is the drag coefficient and

v

is velocity. At terminal velocity, we have:

mg = kv_{terminal}

Therefore:

v_{terminal} = \frac{mg}{k}

Since both balls have the same drag coefficient

k = 0.5 \text{ s}^{-1}

, their terminal velocities are:

Ball 1: $v_1 = \frac{(2)(9.8)}{0.5} = 39.2 \text{ m/s}$
Ball 2: $v_2 = \frac{(8)(9.8)}{0.5} = 156.8 \text{ m/s}$

The ratio is

v_1/v_2 = \frac{2g/k}{8g/k} = \frac{2}{8} = \frac{1}{4}

, confirming answer D. Let's examine why the other options are wrong: A is incorrect because terminal velocity clearly depends on mass in the formula

v_{terminal} = \frac{mg}{k}

. B reverses the relationship—while lighter objects do experience relatively more drag per unit mass, this actually makes them fall slower, not faster. C suggests a square root relationship, but linear drag produces a direct proportionality between terminal velocity and mass. Study tip: Remember that terminal velocity always equals

\frac{mg}{\text{drag coefficient}}

. For linear drag, velocity is directly proportional to mass; for quadratic drag (

F_d = kv^2

), it's proportional to the square root of mass.

Question 15

A particle falls from rest through a medium with linear drag. The velocity as a function of time is $v(t) = 20(1 - e^{-0.3t}) \text{ m/s}$ . At what time does the particle's acceleration equal $3 \text{ m/s}^2$ ?

$t = \frac{1}{0.3}\ln(1.5) \approx 1.35 \text{ s}$ from the velocity-acceleration coupling
$t = \frac{1}{0.3}\ln(3) \approx 3.66 \text{ s}$ using the exponential decay relationship
$t = \frac{1}{0.3}\ln(2) \approx 2.31 \text{ s}$ from direct acceleration analysis (correct answer)
$t = \frac{1}{0.3}\ln(4) \approx 4.62 \text{ s}$ considering the approach to equilibrium

Explanation: When you encounter a particle motion problem with a given velocity function, remember that acceleration is simply the derivative of velocity with respect to time. This connects kinematics directly to calculus. Given

v(t) = 20(1 - e^{-0.3t})

, you find acceleration by taking the derivative:

a(t) = \frac{dv}{dt} = 20 \cdot \frac{d}{dt}(1 - e^{-0.3t}) = 20 \cdot (0 - (-0.3)e^{-0.3t}) = 6e^{-0.3t}

To find when

a(t) = 3 \text{ m/s}^2

, set up the equation:

6e^{-0.3t} = 3

Solving:

e^{-0.3t} = \frac{1}{2}

Taking the natural logarithm:

-0.3t = \ln(\frac{1}{2}) = -\ln(2)

Therefore:

t = \frac{\ln(2)}{0.3} \approx 2.31 \text{ s}

This confirms answer C is correct. A uses

\ln(1.5)

, which would arise from incorrectly setting

6e^{-0.3t} = 4

instead of 3. B uses

\ln(3)

, which comes from mistakenly setting

e^{-0.3t} = \frac{1}{3}

rather than

\frac{1}{2}

. D uses

\ln(4)

, suggesting an error where someone set

e^{-0.3t} = \frac{1}{4}

. Study tip: Always remember that acceleration equals

\frac{dv}{dt}

. When working with exponential functions, double-check your algebra when isolating the exponential term—small errors in fractions lead to completely different logarithmic answers.

Question 16

A skydiver experiences two phases of motion: free fall with minimal drag ( $k_1 \approx 0$ ) followed by descent with significant linear drag ( $k_2 = 0.8 \text{ s}^{-1}$ ) after reaching velocity $v_1 = 60 \text{ m/s}$ . In the second phase, what is the velocity after $2.5$ seconds if the terminal velocity is $50 \text{ m/s}$ ?

$v = 55e^{-2.0} \approx 7.44 \text{ m/s}$ using direct exponential decay model
$v = 50 - 10e^{-2.0} \approx 48.65 \text{ m/s}$ using exponential approach from below
$v = 50 + 10e^{-2.0} \approx 51.35 \text{ m/s}$ using exponential approach from above (correct answer)
$v = 60 - 10e^{-2.0} \approx 58.65 \text{ m/s}$ using initial velocity reference

Explanation: When analyzing motion with air resistance, you need to set up a differential equation that accounts for how drag force affects acceleration. For a falling object with linear drag, the equation is

\frac{dv}{dt} = g - kv

, where terminal velocity occurs when

g = kv_t

. Since terminal velocity is 50 m/s and

k_2 = 0.8 \text{ s}^{-1}

, we can verify:

g = 0.8 \times 50 = 40 \text{ m/s}^2

. The differential equation becomes

\frac{dv}{dt} = 40 - 0.8v

. Rearranging:

\frac{dv}{dt} = -0.8(v - 50)

. This separable equation gives us

v(t) = 50 + (v_0 - 50)e^{-0.8t}

. Since the skydiver enters the second phase at

v_0 = 60 \text{ m/s}

, we have

v(t) = 50 + 10e^{-0.8t}

. After 2.5 seconds:

v = 50 + 10e^{-2.0} \approx 51.35 \text{ m/s}

. This is answer C. Answer A incorrectly models simple exponential decay without considering terminal velocity equilibrium. Answer B assumes the skydiver starts below terminal velocity and approaches from below, contradicting the given initial condition of 60 m/s. Answer D uses the wrong equilibrium point, treating the initial velocity as the target rather than terminal velocity. The key insight is recognizing that when an object starts above terminal velocity, it exponentially approaches that terminal velocity from above. Always identify whether the initial condition is above or below the equilibrium value to determine the direction of approach.

Question 17

A sphere falls through a viscous fluid where the drag force follows $F_d = kv$ . The equation of motion is $m\frac{dv}{dt} = mg - kv$ . If the mass is doubled while keeping the drag coefficient constant, how does the time to reach $95\%$ of terminal velocity change?

The time doubles because the time constant $\tau = m/k$ doubles proportionally (correct answer)
The time halves because increased mass accelerates the approach to equilibrium
The time remains unchanged because terminal velocity scaling compensates exactly
The time increases by $\sqrt{2}$ because mass affects the exponential rate

Explanation: The time constant is

\tau = m/k

. The approach to terminal velocity follows

v(t) = v_t(1 - e^{-t/\tau})

. To reach

95\%

0.95 = 1 - e^{-t/\tau}

, so

t = \tau\ln(20)

. When mass doubles,

\tau

doubles, so the time to reach

95\%

also doubles. The terminal velocity also doubles (

v_t = mg/k

), but this doesn't affect the time scale. Choice B confuses mass with acceleration effects. Choice C ignores the time constant change. Choice D applies incorrect scaling.

Question 18

A falling object with linear drag has the differential equation $\frac{dv}{dt} + kv = g$ . If $k = 0.4 \text{ s}^{-1}$ and $g = 10 \text{ m/s}^2$ , and the object is dropped from rest, what is the acceleration when the velocity reaches half the terminal velocity?

$a = 2.5 \text{ m/s}^2$ because drag force reduces the net acceleration
$a = 10.0 \text{ m/s}^2$ because initial conditions determine acceleration
$a = 7.5 \text{ m/s}^2$ because gravitational dominance remains significant
$a = 5.0 \text{ m/s}^2$ because half-velocity creates half-maximum drag (correct answer)

Explanation: When you encounter a differential equation for motion with drag, remember that acceleration equals the net force divided by mass, which translates directly to the right side of your DE when written in standard form. The given equation

\frac{dv}{dt} + kv = g

can be rewritten as

\frac{dv}{dt} = g - kv

. Since

\frac{dv}{dt}

is acceleration, we have

a = g - kv

. First, find the terminal velocity where acceleration becomes zero:

0 = g - kv_t

, so

v_t = \frac{g}{k} = \frac{10}{0.4} = 25 \text{ m/s}

. When velocity reaches half the terminal velocity:

v = \frac{25}{2} = 12.5 \text{ m/s}

Substituting into the acceleration equation:

a = 10 - 0.4(12.5) = 10 - 5 = 5.0 \text{ m/s}^2

Answer D correctly recognizes that at half terminal velocity, the drag force is exactly half its maximum value, leaving half the gravitational acceleration unbalanced. Answer A uses the wrong fraction—at half velocity, drag reduces acceleration by half of its maximum effect, not to one-quarter of gravitational acceleration. Answer B ignores that drag increases with velocity, incorrectly assuming constant gravitational acceleration throughout the fall. Answer C miscalculates by not properly accounting for the linear relationship between velocity and drag force. Key strategy: For linear drag problems, always remember that

a = g - kv

. Terminal velocity occurs when these balance, so at any fraction of terminal velocity, you can quickly calculate the remaining unbalanced acceleration.

Question 19

An object is thrown vertically upward with an initial speed $v_0$ . It is subject to both gravity and linear air resistance. Which statement is true about its motion?

The time taken to reach the maximum height is the same as the time taken to fall back to the initial position.
The speed of the object when it returns to its initial position is equal to the initial speed $v_0$ .
The magnitude of the object's acceleration is greatest at the peak of its trajectory.
The time taken to reach the maximum height is less than the time it would take in a vacuum. (correct answer)

Explanation: Let downward be the positive direction. The equation of motion is

m(dv/dt) = mg - kv

. On the way up, velocity

v

is negative, so the drag force

-kv

is positive (downward). Thus, drag and gravity both act downward, so the magnitude of acceleration is

|a| = |g - (k/m)v| = g + (k/m)|v| > g

. In a vacuum, the acceleration magnitude is always

g

. Since the upward-moving object with drag has a greater downward acceleration, it will slow down more quickly and reach its maximum height in less time than in a vacuum. A and B are false because air resistance is a non-conservative force that removes mechanical energy from the system. C is false because the acceleration magnitude

g + (k/m)|v|

is greatest when

|v|

is greatest, which occurs at the start of the motion, not at the peak where

v=0