Differential Equations Quiz: Falling Objects With Drag

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.

Question 1

1. $v_T \tau (1 + e^{-2})$ (correct answer)
2. $v_T \tau (2 + e^{-2})$
3. $v_T (1 - e^{-2})$
4. $2 v_T \tau$

Explanation: First, find the velocity function: $v(t) = v_T(1 - e^{-t/\tau})$. To find the distance fallen, integrate the velocity from $t=0$ to $t=2\tau$. 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

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

1. The particle will speed up, with its velocity approaching infinity.
2. The particle will maintain a constant speed of $10$ m/s as there is no net force.
3. The particle will slow down, with its speed approaching $5$ m/s from above. (correct answer)
4. 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

1. $A$ is the initial acceleration $g$, and $B$ is the time constant $m/k$.
2. $A$ is the terminal velocity $mg/k$, and $B$ is the reciprocal of the time constant, $k/m$. (correct answer)
3. $A$ is the terminal velocity $mg/k$, and $B$ is the time constant $m/k$.
4. $A$ 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 $v(0)=0$ is $v(t) = \frac{mg}{k}(1 - e^{-(k/m)t})$. 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

1. $20 - 20e^{-1}$ m/s
2. $20 + 10e^{-1}$ m/s
3. $20 - 10e^{-4}$ m/s
4. $20 - 10e^{-1}$ 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

1. $1$ s (correct answer)
2. $\ln(2)$ s
3. $2$ s
4. $1/\ln(2)$ 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

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

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

1. $A = v_T$, $B = v_T \tau$ (correct answer)
2. $A = v_T$, $B = 0$
3. $A = g$, $B = v_T \tau$
4. $A = v_T$, $B = \tau$

Explanation: To find the distance $x(t)$, we integrate the velocity function $v(t)$ from $0$ to $t$, 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

1. $gt$ (correct answer)
2. $0$
3. $g$
4. The limit is infinite.

Explanation: To find the limit of $v(t)$ as $k \to 0$, 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

1. Her velocity instantaneously drops to the new, lower terminal velocity $mg/k_2$.
2. She immediately begins to decelerate, regardless of her speed $v_1$ when the parachute opened.
3. If her speed $v_1$ was greater than the new terminal velocity $mg/k_2$, she will decelerate. (correct answer)
4. 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

1. The time taken to reach the maximum height is the same as the time taken to fall back to the initial position.
2. The speed of the object when it returns to its initial position is equal to the initial speed $v_0$.
3. The magnitude of the object's acceleration is greatest at the peak of its trajectory.
4. 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 $v=0$.

Question 12

1. $t \approx 3.0 \text{ s}$
2. $t \approx 6.0 \text{ s}$ (correct answer)
3. $t \approx 9.0 \text{ s}$
4. $t \approx 12.0 \text{ s}$

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 13

1. $1:4$ (the sphere in air has 4 times the terminal velocity)
2. $4:1$ (the sphere in air has 4 times the terminal velocity) (correct answer)
3. $1:1$ (both spheres have the same terminal velocity)
4. $2:1$ (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 14

1. $t_2 = 2t_1$ because the exponential relationship doubles the time interval (correct answer)
2. $t_2 = 1.5t_1$ because the velocity increase follows a power law
3. $t_2 = t_1 + \tau$ because each time constant adds the same increment
4. $t_2 = \sqrt{2}t_1$ 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 15

1. $v_1/v_2 = 1:1$ because terminal velocity is independent of mass
2. $v_1/v_2 = 4:1$ because lighter objects experience relatively more drag
3. $v_1/v_2 = 1:2$ because velocity scales with the square root of mass
4. $v_1/v_2 = 1:4$ 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: $v_2 = \frac{(8)(9.8)}{0.5} = 156.8 \text{ m/s}$

Question 16

1. $t = \frac{1}{0.3}\ln(1.5) \approx 1.35 \text{ s}$ from the velocity-acceleration coupling
2. $t = \frac{1}{0.3}\ln(3) \approx 3.66 \text{ s}$ using the exponential decay relationship
3. $t = \frac{1}{0.3}\ln(2) \approx 2.31 \text{ s}$ from direct acceleration analysis (correct answer)
4. $t = \frac{1}{0.3}\ln(4) \approx 4.62 \text{ s}$ 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 17

1. $v = 55e^{-2.0} \approx 7.44 \text{ m/s}$ using direct exponential decay model
2. $v = 50 - 10e^{-2.0} \approx 48.65 \text{ m/s}$ using exponential approach from below
3. $v = 50 + 10e^{-2.0} \approx 51.35 \text{ m/s}$ using exponential approach from above (correct answer)
4. $v = 60 - 10e^{-2.0} \approx 58.65 \text{ m/s}$ 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 18

1. The time doubles because the time constant $\tau = m/k$ doubles proportionally (correct answer)
2. The time halves because increased mass accelerates the approach to equilibrium
3. The time remains unchanged because terminal velocity scaling compensates exactly
4. 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 19

1. $a = 2.5 \text{ m/s}^2$ because drag force reduces the net acceleration
2. $a = 10.0 \text{ m/s}^2$ because initial conditions determine acceleration
3. $a = 7.5 \text{ m/s}^2$ because gravitational dominance remains significant
4. $a = 5.0 \text{ m/s}^2$ 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.