AP Physics C: Mechanics · Topic 2.9

Topic 2.9: Resistive Forces

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

A resistive force is a velocity-dependent force pointing opposite the velocity, and the AP Physics C model is that it equals negative k times the velocity. Put that into Newton's second law and you get a differential equation. Separate the variables, integrate, and the answer is an exponential.

AP Physics: Unit 2 (topics 2.9 Resistive Forces). AP Physics C: Mechanics Unit 2, Topic 2.9. The ONLY topic in this unit with no AP Physics 1 counterpart; AP Physics 1's Topic 2.9 is Circular Motion, which is 2.10 here. One learning objective, 2.9.A, describe the motion of an object subject to a resistive force. Supported by 2.9.A.1 (a resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example F_r = -kv), 2.9.A.2 (applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity), 2.9.A.2.i (using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration), 2.9.A.2.ii (the acceleration or position may be determined using initial conditions and methods of calculus once a function for velocity is determined), 2.9.A.2.iii (position, velocity and acceleration as functions of time for a resistive force OF THE FORM F_r = -kv are exponential and have asymptotes determined by the initial conditions of the object and the forces exerted on the object) and 2.9.A.3 (terminal velocity is the maximum speed achieved by an object moving under the influence of a constant force and a resistive force exerted in opposite directions; the terminal condition is reached when the net force is zero). NO boundary statement. The framework gives ONLY the linear model; there is no quadratic drag law in the required content. The v-squared force in the CED's sample free-response question 2 is an aerodynamic DOWNWARD force on a race car, perpendicular to the velocity, so it is not a resistive force under 2.9.A.1. Verified against the Table of Information appendix rendered as an image: no resistive-force equation, no terminal-velocity expression and no exponential solution is printed on the AP Physics C: Mechanics equation sheet. The Calculus table on that same appendix page does print the integral of dx/(x+a) = ln|x+a|, the integral of e^(ax), and the matching derivatives. The exam-conventions box states that air resistance is assumed to be negligible unless otherwise stated, and the vocabulary appendix adds that drag forces are negligible. Suggested skills are 1.B, 2.A, 2.C, 3.A and 3.C, five in total, the joint largest count in the unit. NONE of the six optional sample instructional activities the CED lists for Unit 2 is on Topic 2.9.

This topic does not exist in AP Physics 1

Topic 2.9 is the only topic in AP Physics C: Mechanics Unit 2 with no counterpart anywhere in AP Physics 1. It is also the reason the two courses number their circular-motion topic differently: AP Physics 1 has nine topics in Unit 2 and puts Circular Motion at 2.9, while AP Physics C inserts Resistive Forces there and pushes Circular Motion to 2.10.

The reason for the gap is structural, not a matter of difficulty. A resistive force depends on the velocity; the velocity depends on the acceleration; the acceleration depends on the force. No substitution unties that loop. The only way out is to accept that Newton's second law has become a differential equation and to solve it, and an algebra-based course has no tool that does that.

The AP Physics C: Mechanics framework says exactly this. Essential knowledge 2.9.A.2 reads: applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity. Statement 2.9.A.2.i names the method: using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.

Nothing else in Unit 2 requires you to solve a differential equation, and the algebra-based course cannot even pose the question.

One more fact worth knowing before you start: the AP Physics C: Mechanics equation sheet prints no resistive-force equation and no exponential solution. Every resistive-force answer is built from Newton's second law and the printed integral rules. That is checked against the Table of Information appendix, not recalled.

What the CED requires of Topic 2.9

One learning objective. Four essential-knowledge statements. No boundary statement.

Objective 2.9.A: describe the motion of an object subject to a resistive force. Suggested skills 1.B, 2.A, 2.C, 3.A and 3.C.

2.9.A.1: a resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example

\vec{F}_r = -k\vec{v}

2.9.A.2: applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
2.9.A.2.i: using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.
2.9.A.2.ii: the acceleration or position of a moving object that is subject to a velocity-dependent force may be determined using initial conditions of the object and methods of calculus, once a function for velocity is determined.
2.9.A.2.iii: the position, velocity, and acceleration as functions of time of an object under the influence of a resistive force of the form $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes that are determined by the initial conditions of the object and the forces exerted on the object.
2.9.A.3: terminal velocity is defined as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions. The terminal condition is reached when the net force exerted on the object is zero.

On the model you are expected to use. The framework prints exactly one resistive-force expression, $\vec{F}_r = -k\vec{v}$ , and statement 2.9.A.2.iii ties its exponential claim specifically to "a resistive force of the form $\vec{F}_r = -k\vec{v}$ ". Nothing in the required content gives a force proportional to $v^2$ .

That matters, because a quadratic model does turn up in the CED's sample free-response section and is easy to misread. Sample question 2 gives race cars for which the air exerts a downward force $F_{\text{down}} = bv^2$ . That force is perpendicular to the velocity, so it is not a resistive force under 2.9.A.1, which requires the force to be in the opposite direction of the object's velocity. It changes the normal force and therefore the friction; it does not decelerate the car directly. Do not carry a quadratic drag law into a Topic 2.9 problem from having seen it.

Statement 2.9.A.2 says the second law "results in a differential equation", with no restriction on which one, so a question could in principle hand you another velocity dependence to separate. What the framework guarantees to be exponential is only the linear case.

The one derivation this topic is built on

Every resistive-force problem in the course is this derivation with different constants. Learn the shape, not the results.

Case 1: resistive force only. An object coasting with nothing else acting along its direction of motion. Newton's second law gives

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

Separate: put every $v$ on one side and every $t$ on the other, $\dfrac{dv}{v} = -\dfrac{k}{m}\, dt$ . Now integrate between the actual limits, which is what 2.9.A.2.i means by "the proper limits of integration": from $v_0$ at $t = 0$ to $v$ at time $t$ . The left side uses $\int \frac{dx}{x+a} = \ln|x+a|$ with $a = 0$ , printed in the Calculus table of the AP Physics C: Mechanics Table of Information. The result is $\ln(v/v_0) = -kt/m$ , so

v(t) = v_0 e^{-t/\tau}, \qquad \tau = \frac{m}{k}

The object slows exponentially and never formally stops. The asymptote is $v = 0$ , set by the fact that no other force acts.

Case 2: a constant force plus a resistive force. An object falling under gravity through air, or being driven at constant thrust against drag. With down positive,

m\frac{dv}{dt} = mg - kv

The terminal speed is where the bracket vanishes, which is 2.9.A.3's condition that the net force is zero: $v_T = mg/k$ . Rewrite the equation using it, $\dfrac{dv}{dt} = \dfrac{v_T - v}{\tau}$ , separate, and integrate from $v_0$ to $v$ . The integral is now literally the printed one, $\int \frac{dv}{v - v_T} = \ln|v - v_T|$ , giving

v(t) = v_T + (v_0 - v_T)e^{-t/\tau}

Read that expression rather than memorising it. It is the asymptote plus the initial gap, and the gap decays with time constant $\tau = m/k$ . Released from rest, $v_0 = 0$ and it becomes $v(t) = v_T(1 - e^{-t/\tau})$ , approaching $v_T$ from below. Thrown down faster than terminal, $v_0 > v_T$ and the same formula approaches $v_T$ from above. That is 2.9.A.2.iii's "asymptotes that are determined by the initial conditions of the object and the forces exerted on the object" in one line: the forces set $v_T$ , the initial conditions set the gap.

Then acceleration and position, per 2.9.A.2.ii. Differentiate the velocity for the acceleration, integrate it for the position, using $\int e^{ax}\, dx = \frac{1}{a}e^{ax}$ from the same Calculus table. Both are exponentials with the same $\tau$ .

a(t) = -\frac{v_0 - v_T}{\tau}e^{-t/\tau}, \qquad x(t) = v_T t + (v_0 - v_T)\tau\left(1 - e^{-t/\tau}\right)

What the sheet prints, and what it leaves to you

There is no resistive-force equation on the AP Physics C: Mechanics equation sheet. The Mechanics table on the Table of Information appendix runs from the kinematic equations through gravitation, friction, springs, circular motion, energy, momentum and rotation, and $\vec{F}_r = -k\vec{v}$ appears nowhere in it. Neither does $v_T = mg/k$ , nor $\tau = m/k$ , nor any exponential solution. That changes how you prepare: there is nothing to look up, so the derivation has to be automatic.

What the booklet does print, in the Calculus table on the same appendix page, is every integral you need:

Printed rule	Where you use it
$\int \dfrac{dx}{x+a} = \ln\lvert x + a\rvert$	separating variables in $dv/(v - v_T)$
$\int e^{ax}\, dx = \dfrac{1}{a}e^{ax}$	integrating $v(t)$ to get $x(t)$
$\dfrac{d}{dx}\left(e^{ax}\right) = ae^{ax}$	differentiating $v(t)$ to get $a(t)$
$\dfrac{d}{dx}\left(\ln ax\right) = \dfrac{1}{x}$	checking the separation step
$\int x^n\, dx = \dfrac{1}{n+1}x^{n+1},\ n \neq -1$	everything except the logarithm

That Calculus table is easy to forget exists, because it is not part of the mechanics equation table and is not what people mean when they say "the equation sheet". It sits on the same appendix page as the geometry, trigonometry, vector and identity boxes. The logarithm rule in the first row is precisely the integral separation of variables produces here, so the mathematics is provided even though the physics is not.

Two other printed lines matter: $\Delta x = \int v_x(t)\, dt$ and $\Delta v_x = \int a_x(t)\, dt$ , the general licence to integrate a velocity function, and $\vec{F}_{\text{net}} = d\vec{p}/dt$ , the form of the second law that makes writing a differential equation the natural first move.

One more absence worth naming, because it changes the default assumption in every other topic. The exam-conventions box on the same appendix page states: "Air resistance is assumed to be negligible unless otherwise stated." The appendix "Vocabulary and Definitions of Important Ideas in AP Physics" repeats it and adds "Drag forces are negligible" to the same list of assumptions. Topic 2.9 is the "otherwise stated" case. If a question does not raise the subject, there is no resistive force.

Terminal velocity is not a property of the object

Statement 2.9.A.3 defines terminal velocity as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force exerted in opposite directions, and says the terminal condition is reached when the net force on the object is zero.

Read the definition carefully and three things follow that students routinely get wrong.

It depends on the driving force, not just the object. For $\vec{F}_r = -k\vec{v}$ the condition $F = kv_T$ gives $v_T = F/k$ . Double the constant force and the terminal speed doubles. A skydiver's terminal speed is not a number attached to the skydiver; it is set jointly by their weight and their drag constant, which is why changing posture changes it.

It needs a constant force to exist at all. Case 1 above, with no driving force, has no terminal velocity. The object simply decays toward zero. Terminal velocity is a feature of the two-force problem.

It is never actually reached. The approach is exponential, so $v = v_T$ only in the limit. After one time constant the gap has shrunk to $e^{-1} = 0.368$ of its original size, after two to $0.135$ , after three to $0.0498$ .

The acceleration falls as the speed rises. Rearranging, $a = g - v/\tau$ for the falling case. At release the acceleration is $g$ ; as $v$ climbs, $a$ shrinks; at $v = v_T$ it is zero. An object approaching terminal velocity is still speeding up the whole time, just by less and less. Saying it "decelerates" is wrong, and saying it "stops accelerating" is only true in the limit.

The terminal velocity glossary entry and the drag force glossary entry give one-paragraph versions of these definitions.

Graphs, experiments, and the linearisation the exam wants

Two of the five suggested skills for Topic 2.9 are experimental: 1.B, create quantitative graphs with appropriate scales and units, including plotting data, and 3.A, create experimental procedures that are appropriate for a given scientific question. So a resistive-force question can arrive as a data-analysis problem rather than as a derivation.

The shapes to recognise.

$v$ against $t$ , released from rest: rises from zero, concave down, flattening toward the horizontal asymptote $v_T$ , with initial slope $g$ .
$v$ against $t$ , launched faster than terminal: falls toward the same asymptote from above, concave up.
$v$ against $t$ , coasting with no driving force: exponential decay toward zero.
$a$ against $t$ : an exponential decay toward zero in all three cases, because $a \propto (v_T - v)$ .
$x$ against $t$ : curved at first, then a straight line of slope $v_T$ . In the coasting case it flattens to a horizontal asymptote at the total distance $v_0\tau$ , finite even though the object never stops.

The linearisation. An exponential is hard to read off a graph and easy to read off a logarithm. For an object released from rest, $v = v_T(1 - e^{-t/\tau})$ rearranges to

\ln\left(1 - \frac{v}{v_T}\right) = -\frac{t}{\tau} = -\frac{k}{m}t

So plotting $\ln(1 - v/v_T)$ against $t$ gives a straight line through the origin with slope $-k/m$ , and the drag constant is the mass times the magnitude of that slope. For the coasting case the trick is simpler still: $\ln v$ against $t$ is a straight line of slope $-k/m$ and intercept $\ln v_0$ . The third worked example runs this on a data set. Measure the terminal speed from the flat part of the graph, form the logarithm, plot, and take the slope. Do not fit a curve by eye.

There is no AP Physics 1 version of this page

AP Physics 1 has no Topic 2.9 Resistive Forces and no equivalent objective anywhere in its framework. Its Topic 2.9 is Circular Motion, which in AP Physics C: Mechanics is Topic 2.10. If you arrived here looking for the algebra-based treatment of air resistance, there is not one to link to: the AP Physics 1 exam conventions assume air resistance is negligible and the course never lifts that assumption.

The nearest thing in the algebra-based course is qualitative: AP Physics 1 students meet terminal velocity as a described phenomenon, but are never asked to produce $v(t)$ .

	AP Physics 1	AP Physics C: Mechanics
Topic on resistive forces	none	2.9, one learning objective
Drag model given	none	$\vec{F}_r = -k\vec{v}$
Terminal velocity	phenomenon only	defined at 2.9.A.3, computed
Method	not applicable	separation of variables
Topic 2.9 in that course	Circular Motion	Resistive Forces
Topic 2.10 in that course	does not exist	Circular Motion

That numbering shift is worth carrying in your head, because search results and textbook indexes mix the two courses freely. "Topic 2.9" means different things depending on which exam you are sitting.

How Topic 2.9 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.9 is one of ten topics in it and has one of the unit's nineteen learning objectives.

Its five suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Five suggested skills is the joint largest count in Unit 2, matched only by Topics 2.5 and 2.7. Skill 2.A alone carries 25 to 30 percent of the multiple-choice section, the largest band the CED lists for any single skill, and the CED lists 2.A among the suggested skills for exactly four topics in this unit: 2.6, 2.7, 2.9 and 2.10.

The free-response question type this topic fits best is question 1, Mathematical Routines, worth 10 points with a suggested time of 20 to 25 minutes, which the CED describes as expecting students to symbolically derive relationships between variables as well as calculate numerical values. Question 3, Experimental Design and Analysis, also fits, given skills 1.B and 3.A.

One thing to notice about how the CED itself treats this topic: none of the six optional sample instructional activities it lists for Unit 2 is on Topic 2.9. They cover Topics 2.1 twice, 2.3, 2.5, 2.7 and 2.10. A course that follows the suggested activities and runs short of time can reach the exam having skimmed the one topic in the unit AP Physics 1 does not cover at all. Budget your own time accordingly.

The Unit 2 hub sets this topic against the other nine, and Topic 2.5 develops the form of the second law you start from.

Coasting to rest: resistive force alone

A small boat of mass $m = 220$ kg is moving at $v_0 = 3.6$ m/s when its engine is switched off. The water exerts a resistive force $\vec{F}_r = -k\vec{v}$ with $k = 44$ kg/s, and no other horizontal force acts. Find (a) the velocity as a function of time, (b) the speed and position after one and after two time constants, and (c) the total distance the boat travels.

Declare the convention: positive in the direction of the initial motion. The resistive force is then negative, as 2.9.A.1 requires, because it points opposite the velocity.
(a) Write Newton's second law with the only horizontal force present. This is the step 2.9.A.2 describes, and it is the step worth the point on a free-response question: $m\dfrac{dv}{dt} = -kv$ .
Separate the variables, per 2.9.A.2.i: $\dfrac{dv}{v} = -\dfrac{k}{m}\, dt$ . Then integrate over the proper limits, from $v_0$ at $t = 0$ to $v$ at time $t$ , using the printed rule $\int \frac{dx}{x+a} = \ln\lvert x+a\rvert$ with $a = 0$ : $\ln v - \ln v_0 = -\dfrac{k}{m}t$ .
Exponentiate: $v(t) = v_0 e^{-kt/m} = v_0 e^{-t/\tau}$ with $\tau = \dfrac{m}{k} = \dfrac{220}{44} = 5.0$ s.
Check the two limits before using it. At $t = 0$ it gives $v_0$ , correct. As $t \to \infty$ it goes to zero, which is the asymptote 2.9.A.2.iii predicts for this set of forces.
(b) At $t = \tau = 5.0$ s: $v = 3.6\,e^{-1} = 3.6(0.36788) = 1.32$ m/s. At $t = 2\tau = 10$ s: $v = 3.6\,e^{-2} = 3.6(0.13534) = 0.487$ m/s.
For position, integrate the velocity, per 2.9.A.2.ii and the printed rule $\int e^{ax} dx = \frac{1}{a}e^{ax}$ : $x(t) = \displaystyle\int_0^t v_0 e^{-t'/\tau}\, dt' = v_0\tau\left(1 - e^{-t/\tau}\right)$ .
Numerically, $v_0\tau = 3.6(5.0) = 18.0$ m. At $t = \tau$ : $x = 18.0(1 - 0.36788) = 18.0(0.63212) = 11.4$ m. At $t = 2\tau$ : $x = 18.0(1 - 0.13534) = 18.0(0.86466) = 15.6$ m.
(c) The total distance is the limit of $x(t)$ as $t \to \infty$ , which is $v_0\tau = \dfrac{mv_0}{k} = 18.0$ m. The boat never formally stops, and yet it travels a finite distance. That pairing is the least intuitive result in this topic, and it is worth checking rather than believing.
Sanity check the distance a second way. The impulse of the resistive force must remove all the momentum: $\int F\, dt = -k\int v\, dt = -k \times (\text{total distance})$ , and that must equal $0 - mv_0$ . So the distance is $mv_0/k = 220(3.6)/44 = 18.0$ m, which matches. The initial acceleration is $a_0 = -v_0/\tau = -0.72\ \mathrm{m/s^2}$ , largest at the start and decaying with the same time constant.

(a) $v(t) = v_0 e^{-t/\tau}$ with $\tau = m/k = 5.0$ s. (b) $v = 1.32$ m/s and $x = 11.4$ m at $t = 5.0$ s; $v = 0.487$ m/s and $x = 15.6$ m at $t = 10$ s. (c) The total distance is $mv_0/k = 18.0$ m, reached only in the limit, since the speed never becomes exactly zero.

Falling faster than terminal velocity

A ball of mass $m = 0.060$ kg is thrown straight down with an initial speed $v_0 = 8.0$ m/s. The air exerts a resistive force $\vec{F}_r = -k\vec{v}$ with $k = 0.24$ kg/s. Using $g = 9.8\ \mathrm{m/s^2}$ , find the terminal speed, the initial acceleration, the speed after one and after three time constants, and the distance fallen in three time constants.

Declare the convention: down is positive. Gravity contributes $+mg$ and the resistive force $-kv$ , since the ball is moving downward.
Newton's second law: $m\dfrac{dv}{dt} = mg - kv$ . Two forces in opposite directions with one of them velocity-dependent is exactly the configuration 2.9.A.3 describes, so a terminal velocity exists here.
Terminal speed from 2.9.A.3's condition that the net force is zero: $mg = kv_T$ , so $v_T = \dfrac{mg}{k} = \dfrac{(0.060)(9.8)}{0.24} = \dfrac{0.588}{0.24} = 2.45$ m/s.
The initial speed is 8.0 m/s, well above $v_T$ . So the ball must slow down toward 2.45 m/s rather than speed up toward it. Predicting that before you solve anything is the check that you have read 2.9.A.2.iii correctly: the asymptote comes from the forces, the direction of approach from the initial conditions.
Initial acceleration, straight from the second law: $a_0 = g - \dfrac{kv_0}{m} = 9.8 - \dfrac{(0.24)(8.0)}{0.060} = 9.8 - 32.0 = -22.2\ \mathrm{m/s^2}$ . Negative in the down-positive convention, so it points upward: at release the drag is more than three times the weight.
Solve. Write the equation using $v_T$ and $\tau = m/k = 0.060/0.24 = 0.25$ s: $\dfrac{dv}{dt} = \dfrac{v_T - v}{\tau}$ . Separating gives $\dfrac{dv}{v - v_T} = -\dfrac{dt}{\tau}$ , and the left side is the printed integral $\int \frac{dx}{x+a} = \ln\lvert x+a\rvert$ with $a = -v_T$ .
Integrate from $v_0$ at $t = 0$ to $v$ at time $t$ : $\ln\lvert v - v_T\rvert - \ln\lvert v_0 - v_T\rvert = -t/\tau$ , so $v(t) = v_T + (v_0 - v_T)e^{-t/\tau}$ . Here $v_0 - v_T = 8.0 - 2.45 = 5.55$ m/s, a positive gap, confirming the approach is from above.
At $t = \tau = 0.25$ s: $v = 2.45 + 5.55(0.36788) = 2.45 + 2.042 = 4.49$ m/s.
At $t = 3\tau = 0.75$ s: $v = 2.45 + 5.55(0.049787) = 2.45 + 0.276 = 2.73$ m/s. The gap has shrunk from 5.55 m/s to 0.28 m/s, 5.0 percent of its original size, as three time constants should give.
Distance fallen, integrating the velocity per 2.9.A.2.ii: $x(t) = v_T t + (v_0 - v_T)\tau\left(1 - e^{-t/\tau}\right)$ . At $t = 0.75$ s that is $2.45(0.75) + 5.55(0.25)(1 - 0.049787) = 1.8375 + 1.3875(0.95021) = 1.8375 + 1.3184 = 3.16$ m.
Compare with free fall over the same interval, $v_0 t + \frac{1}{2}gt^2 = 6.0 + 2.76 = 8.76$ m. Resistance has cut the distance by nearly two thirds, which is the size of effect the exam's default assumption of negligible air resistance hides everywhere else in the course.

$v_T = mg/k = 2.45$ m/s and $\tau = m/k = 0.25$ s. The initial acceleration is $22.2\ \mathrm{m/s^2}$ directed upward. $v(t) = v_T + (v_0 - v_T)e^{-t/\tau}$ gives 4.49 m/s at $t = 0.25$ s and 2.73 m/s at $t = 0.75$ s, and the ball falls 3.16 m in those 0.75 s, against 8.76 m in free fall.

Getting the drag constant out of data

A $0.40$ kg object is released from rest and falls through a fluid. Its speed is measured at four times, and the terminal speed is measured as $4.9$ m/s. The data are $t = 0.25$ s, $v = 1.93$ m/s; $t = 0.50$ s, $v = 3.10$ m/s; $t = 0.75$ s, $v = 3.81$ m/s; $t = 1.00$ s, $v = 4.24$ m/s. Determine the drag constant $k$ , and check the result against the terminal speed.

Decide what to plot before touching the data. The raw $v$ against $t$ graph is an exponential approach, and reading a constant off a curve by eye is not a method. Skill 1.B asks for a quantitative graph, which here means a linearised one.
Derive the linearisation. Released from rest, $v_0 = 0$ , so $v(t) = v_T\left(1 - e^{-t/\tau}\right)$ . Rearranging, $1 - \dfrac{v}{v_T} = e^{-t/\tau}$ , and taking logarithms gives $\ln\left(1 - \dfrac{v}{v_T}\right) = -\dfrac{t}{\tau} = -\dfrac{k}{m}t$ .
So a plot of $\ln(1 - v/v_T)$ on the vertical axis against $t$ on the horizontal axis should be a straight line through the origin with slope $-k/m$ , in units of inverse seconds.
Compute the vertical coordinate for each point, using $v_T = 4.9$ m/s. At $t = 0.25$ : $1 - 1.93/4.9 = 0.6061$ , and $\ln(0.6061) = -0.50$ .
At $t = 0.50$ : $1 - 3.10/4.9 = 0.3673$ , and $\ln(0.3673) = -1.00$ .
At $t = 0.75$ : $1 - 3.81/4.9 = 0.2224$ , and $\ln(0.2224) = -1.50$ .
At $t = 1.00$ : $1 - 4.24/4.9 = 0.1347$ , and $\ln(0.1347) = -2.00$ .
Take the slope from the first and last points: $\dfrac{-2.00 - (-0.50)}{1.00 - 0.25} = \dfrac{-1.50}{0.75} = -2.0\ \mathrm{s^{-1}}$ . The intermediate points fall on the same line, which is the evidence that the model is right; a curved plot here would mean the resistive force is not proportional to $v$ .
Convert the slope to the constant: slope $= -k/m$ , so $k = -m \times \text{slope} = (0.40)(2.0) = 0.80$ kg/s. The time constant is $\tau = m/k = 0.50$ s, which is also just the negative reciprocal of the slope.
Check against the terminal speed, which was measured independently. Statement 2.9.A.3 gives $v_T = mg/k = (0.40)(9.8)/0.80 = 3.92/0.80 = 4.9$ m/s. That matches the measured value, so the two determinations of $k$ agree.
Note what the check is worth: the slope used only the shape of the approach, while the terminal speed used only the balance of forces at the end. Two independent routes to the same $k$ is the kind of consistency argument skill 3.C asks for.
A procedural point for skill 3.A: the terminal speed must be read from the flat part of the curve, well past a few time constants, and its uncertainty propagates into every plotted point through the ratio $v/v_T$ . A terminal speed read too early makes the last few points bend away from the line.

Plotting $\ln(1 - v/v_T)$ against $t$ gives a straight line of slope $-2.0\ \mathrm{s^{-1}}$ , so $k = -m \times \text{slope} = 0.80$ kg/s and $\tau = 0.50$ s. The independent check $v_T = mg/k = 4.9$ m/s reproduces the measured terminal speed.

Frequently asked questions

What is a resistive force in AP Physics C?

Essential knowledge 2.9.A.1 of the AP Physics C: Mechanics course and exam description defines a resistive force as a velocity-dependent force in the opposite direction of an object's velocity, and gives as its example that the resistive force equals negative k times the velocity. Two conditions must hold: the force depends on the velocity, and it points opposite to it. A force that depends on speed but points somewhere else, such as an aerodynamic downforce on a car, is not a resistive force under this definition even though it is velocity-dependent.

How do you solve a resistive force problem with separation of variables?

Write Newton's second law with every force present, which gives a differential equation for the velocity, as essential knowledge 2.9.A.2 states. Then rearrange so that every appearance of v is on one side with dv and every t is on the other with dt, and integrate both sides over the proper limits of integration, which is the phrase essential knowledge 2.9.A.2.i uses: from the initial velocity at time zero to the velocity v at time t. For the falling case the velocity-side integral is dv over v minus the terminal speed, which is the printed rule that the integral of dx over x plus a equals the natural log of the absolute value of x plus a. Exponentiating gives the velocity function; differentiate it for acceleration or integrate it for position.

Is there a quadratic drag force in the AP Physics C CED?

Not in the required course content for Topic 2.9. The framework prints exactly one resistive-force expression, the resistive force equals negative k times the velocity, and essential knowledge 2.9.A.2.iii ties its claim that the motion is exponential specifically to a resistive force of that form. A velocity-squared force does appear in the CED's sample free-response question 2, but it is an aerodynamic downward force on a race car, perpendicular to the velocity rather than opposite it, so it is not a resistive force under the 2.9.A.1 definition. It affects the normal force and hence the friction, not the speed directly.

Why is terminal velocity never actually reached?

Because the approach is exponential. Essential knowledge 2.9.A.2.iii says the position, velocity and acceleration of an object under a resistive force of the form negative k times velocity are exponential and have asymptotes determined by the initial conditions and the forces. The velocity is the terminal speed plus the initial gap times e to the power of negative t over the time constant, and that second term shrinks toward zero without ever equalling it. After one time constant the gap is about 37 percent of its starting size, after two about 14 percent, after three about 5 percent. In practice a few time constants is indistinguishable from terminal; mathematically it is never exact.

Does terminal velocity depend on the object or on the forces?

On both, and that is the point of the definition. Essential knowledge 2.9.A.3 defines terminal velocity as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force exerted in opposite directions, with the terminal condition reached when the net force is zero. Setting the two equal for a linear resistive force gives the terminal speed as the constant force divided by k, so doubling the driving force doubles it. It is not a fixed property of the object alone, which is why a skydiver's terminal speed changes with their posture.

Is the resistive force equation on the AP Physics C Mechanics equation sheet?

No. The Mechanics table in the AP Physics C: Mechanics Table of Information prints no resistive-force equation, no terminal-velocity expression and no exponential solution. What it does print, on the same appendix page in the Calculus table, is every rule the derivation needs: the integral of dx over x plus a equals the natural log of the absolute value of x plus a, the integral of e to the ax equals one over a times e to the ax, and the matching derivatives. The mathematics is supplied and the physics is not.

How do you find the drag constant from experimental data?

Linearise, then take a slope. For an object released from rest with a resistive force proportional to velocity, the speed is the terminal speed times one minus e to the negative t over tau. Rearranging and taking logarithms gives that the natural log of one minus v over the terminal speed equals negative k over m times t. Plot that logarithm against time and you should get a straight line through the origin of slope negative k over m, so k is the mass times the magnitude of the slope. If the points curve, the resistive force is not proportional to the first power of the velocity. Cross-check against the terminal speed, which independently gives k as mg divided by the terminal speed.