3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A ball is thrown straight up, and air resistance on it is not negligible. The diagram shows the ball's velocity at two instants, P and Q. Which describes the direction of the resistive force exerted on the ball by the air at P and at Q?
Answer and reasoning
AUpward at P and upward at Q A student who thinks air resistance always opposes gravity picks this. It opposes the velocity: while the ball rises, the air pushes it down, adding to gravity's effect.
BDownward at P and upward at QCorrect A resistive force points opposite to the object's velocity. At P the ball moves up, so the force points down; at Q the ball moves down, so the force points up.
CZero at P and upward at Q A student who thinks air resistance acts only on falling objects picks this. The ball moves through the air at P too, so a resistive force acts, opposite to its upward velocity.
DDownward at P and downward at Q A student who gives the resistive force the direction of the net force picks this. The net force is downward at both instants, but at Q the resistive force alone points up, against the downward velocity.
A ball of mass m is thrown straight up and experiences a resistive force F⃗r = −kv⃗. Upward is taken as positive, and g is the magnitude of the acceleration due to gravity. Which differential equation describes the ball's velocity v while it is rising?
Answer and reasoning
Adv/dt = −g + (k/m)v A student who takes air resistance to point up, opposing gravity, picks this. While the ball rises its velocity is upward, so the resistive force, −kv, points down.
Bdv/dt = +g − (k/m)v A student who reuses the falling-object equation, written with downward positive, picks this. With upward positive, gravity's term is −g.
Cdv/dt = +g + (k/m)v A student who adds the sizes of the two forces, mg + kv, without signs for their directions picks this. With upward positive, both forces point down while the ball rises, so both terms are negative.
Ddv/dt = −g − (k/m)vCorrect Newton's second law with upward positive: gravity contributes −mg, and the resistive force −kv points down while v is positive. So m dv/dt = −mg − kv, and dividing by m gives dv/dt = −g − (k/m)v.
Working Up positive: ΣF = −mg + (−kv) = m dv/dt → dv/dt = −g − (k/m)v. Units: k/m in 1/s, so (k/m)v in m/s².
A skydiver falls at her terminal velocity. Which statement about the skydiver is correct?
Answer and reasoning
AGravity is slightly larger than the resistive force, keeping her moving. A student who thinks motion needs a net force in its direction picks this. Any unbalanced downward force would make her speed up; at constant velocity the forces balance.
BAir resistance balances her weight, so the net force on her is zero.Correct Terminal velocity is reached when the net force is zero: the upward resistive force equals her weight in magnitude, so her velocity stays constant.
CNo forces act on her now, because her velocity is no longer changing. A student who reads 'zero net force' as 'no forces' picks this. Earth still pulls her down and the air still pushes her up; the two forces add to zero.
DHer acceleration is at its maximum, since her speed is at its maximum. A student who links large acceleration with large speed picks this. Her speed is constant, so her acceleration is zero; it was largest, g, at the start of the fall.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.9.A.1 Resistive force, F⃗rFix
Resistive force, F⃗r
A velocity-dependent force exerted on an object in the direction opposite to its velocity, such as the drag of air or water. In the linear model F⃗r = −kv⃗. Unit: newton (N).
Resistive constant, k
The constant of proportionality in F⃗r = −kv⃗; it depends on the object's size and shape and on the fluid. Unit: kilogram per second (kg/s), since N/(m/s) = kg/s.
Students often think Air resistance always acts upward, opposing gravity. In fact No. A resistive force points opposite to the object's velocity. On a rising ball it points down; on a falling ball it points up.
Students often think Air resistance acts only on objects that are falling. In fact No. A resistive force acts on any object moving through a fluid, whether it rises, falls or moves sideways.
2.9.A.2 Differential equation for velocity Fix
Differential equation for velocity
Newton's second law with a resistive force gives an equation relating dv/dt to v, for example m dv/dt = mg − kv for an object falling (downward positive) or m dv/dt = −kv for an object coasting horizontally.
Separation of variables
A method of solving such an equation: collect every v on one side and t on the other, e.g. dv/v = −(k/m)dt, and integrate each side between matching limits, from the initial velocity v₀ at t = 0 to v at time t.
Position and acceleration from v(t)
Once v(t) is known, the acceleration is a = dv/dt and the position is x(t) = x₀ + ∫₀ᵗ v dt, using the object's initial conditions.
Exponential approach and time constant
For F⃗r = −kv⃗, velocity, acceleration and position change exponentially with time, with exponent −kt/m. The time m/k (in s) sets how quickly they approach their asymptotes; after each interval m/k the remaining difference from the asymptote falls to e⁻¹ ≈ 0.37 of its value.
Asymptote
The value a quantity approaches but does not reach in finite time. Its value is set by the initial conditions and the forces: a coasting object's speed approaches zero and its position approaches x₀ + mv₀/k; a falling object's speed approaches mg/k.
Students often think m dv/dt = mg − kv describes any object moving vertically with a resistive force, whatever direction is taken as positive. In fact No. The signs in Newton's second law depend on the chosen positive direction and on the direction of each force at that moment. m dv/dt = mg − kv holds with downward positive; with upward positive the equation is m dv/dt = −mg − kv.
Students often think The time constant is k/m (equivalently, the exponent is −mt/k). In fact No. Separating variables gives the exponent −kt/m, so the time constant is m/k, which has units of seconds; k/m has units of 1/s.
2.9.A.3 Terminal velocity Fix
Terminal velocity
The maximum speed reached by an object acted on by a constant force and a resistive force in opposite directions, reached when the net force is zero. For a falling object with F⃗r = −kv⃗, vT = mg/k.
Students often think All objects fall in the same way, so objects with the same resistive constant reach the same terminal speed whatever their masses. In fact No. For F⃗r = −kv⃗, vT = mg/k: with the same k, a heavier object reaches a larger terminal speed.
Students often think Terminal speed is proportional to the square root of the weight (or the resistive constant goes as 1/vT²) whatever the resistive-force model. In fact Only for a resistive force proportional to v². For the linear model F⃗r = −kv⃗, vT = mg/k is proportional to the weight.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
A 0.40 kg toy boat glides across still water in a straight line with its motor off. The water exerts a resistive force F⃗r = −kv⃗ with k = 0.50 kg/s, and no other horizontal force acts. How long does it take the boat's speed to fall to half its initial value?
Answer and reasoning
A0.55 sCorrect Separating variables in m dv/dt = −kv gives ∫dv/v = −(k/m)∫dt, so v = v₀e−kt/m. The speed halves when kt/m = ln 2: t = (m/k) ln 2 = (0.80 s)(0.693) = 0.55 s.
B0.40 s A student who keeps the deceleration at its initial value, kv₀/m, picks this: the speed would halve at m/(2k) = 0.40 s. The force shrinks as the boat slows, so halving takes longer.
C0.80 s A student who takes the time constant m/k = 0.80 s to be the time for the speed to halve picks this. At t = m/k the speed is e⁻¹ ≈ 0.37 of its initial value; it fell to one half earlier, at (m/k) ln 2.
D0.87 s A student who uses k/m as the time constant picks this: (1.25)(0.693) = 0.87. k/m has units of 1/s, not s; the exponent in the solution is −kt/m.
Working m dv/dt = −kv → ∫v₀v dv/v = −(k/m)∫0t dt → v = v₀e−kt/m. v = v₀/2 at t = (m/k) ln 2 = (0.40 kg/0.50 kg/s)(0.693) = 0.55 s.
A boat of mass m coasts in a straight line with its engine off. At t = 0 its speed is v₀, and the water exerts a resistive force F⃗r = −kv⃗; no other horizontal force acts. How far does the boat travel between t = 0 and t = m/k?
Answer and reasoning
A1.00 mv₀/k A student who thinks the motion is complete after one time constant picks this, the total distance, mv₀/k. At t = m/k the boat still moves at 0.37v₀, and it approaches mv₀/k only as t grows without limit.
B0.37 mv₀/k A student who takes the fraction of distance covered to equal the fraction of speed remaining picks this. e⁻¹ is the fraction of the speed left; the distance covered is (1 − e⁻¹) of the total.
C0.63 mv₀/kCorrect Separating variables gives v = v₀e−kt/m. Integrating, x = ∫₀ᵗ v₀e−kt′/m dt′ = (mv₀/k)(1 − e−kt/m). At t = m/k, x = (1 − e⁻¹)mv₀/k ≈ 0.63 mv₀/k: the boat has covered about 63 percent of the total distance it will ever cover.
D0.50 mv₀/k A student who treats the deceleration as constant at kv₀/m picks this: the boat would stop at t = m/k after v₀t − (1/2)(kv₀/m)t² = 0.50 mv₀/k. The deceleration falls as the boat slows, so it travels farther.
Working v = v₀e−kt/m; x(t) = ∫₀ᵗ v dt = (mv₀/k)(1 − e−kt/m). At t = m/k: x = (1 − e⁻¹)(mv₀/k) = 0.632 mv₀/k.
An object is dropped from rest and falls, experiencing a resistive force F⃗r = −kv⃗. The graph shows how one quantity Q describing the object's motion changes with time t after release. Which quantity could Q be?
Answer and reasoning
AThe size of its acceleration A student who thinks acceleration is largest when speed is largest picks this. The acceleration is g at release, when the speed is zero, and decreases toward zero as the speed approaches its terminal value.
BThe distance it has fallen A student who expects the distance to level off when the speed does picks this. The object keeps falling at nearly its terminal speed, so the distance keeps growing almost linearly with no horizontal asymptote.
CThe net force exerted on it A student who thinks a faster-moving object needs a larger net force picks this. The net force, mg − kv, is largest at release and falls toward zero as the speed rises.
DThe magnitude of its velocityCorrect The speed starts at zero and rises, but ever more slowly, because the growing resistive force reduces the net force. It approaches the terminal speed mg/k, the horizontal asymptote.
Two balls of the same size and shape, so with the same resistive constant k, have masses m and 4m. Each falls with a resistive force F⃗r = −kv⃗. What is the ratio of the terminal speed of the 4m ball to that of the m ball?
Answer and reasoning
A1.00 A student who extends 'all objects fall alike' to falls with air resistance picks this. That holds only when the resistive force is negligible; here the heavier ball needs a larger resistive force, and so a larger speed, to balance its weight.
B2.00 A student who uses the quadratic-drag result, vT ∝ √(mg), picks this: √4 = 2. For the linear model F⃗r = −kv⃗, vT = mg/k is proportional to the mass.
C4.00Correct At terminal velocity the net force is zero: kvT = mg, so vT = mg/k. With the same k, vT is proportional to the mass, and the ratio is 4m/m = 4.00.
D0.25 A student who thinks the heavier ball has more air resistance and so falls more slowly picks this. The resistive force depends on speed, not mass; the heavier ball must reach a higher speed before kv equals its weight.
Working Terminal condition: kvT − mg = 0 → vT = mg/k. Ratio (4m)g/k ÷ mg/k = 4.00.
A ball of mass m is thrown straight down with speed 2vT, where vT = mg/k is its terminal speed for the resistive force F⃗r = −kv⃗ it experiences. What is its speed at time t = (m/k) ln 2 after it is thrown?
Answer and reasoning
A1.5vTCorrect With downward positive, m dv/dt = mg − kv. Separating variables from v = 2vT at t = 0 gives v = vT + (2vT − vT)e−kt/m. At t = (m/k) ln 2, e−kt/m = 1/2, so v = 1.5vT: the ball slows toward vT.
B0.5vT A student who uses the from-rest solution, vT(1 − e−kt/m), whatever the initial speed picks this. That solution gives 0 at t = 0, not 2vT; the initial condition fixes the solution.
C1.0vT A student who expects the resistive force to bring the ball to rest uses 2vT e−kt/m, which gives 1.0vT. Gravity also acts, so the speed approaches vT, not zero.
D3.0vT A student who expects a downward-moving ball always to speed up writes a growing solution, vT(1 + ekt/m), which gives 3.0vT. Above vT the resistive force exceeds the weight, so the ball slows down.
Working Down positive: dv/dt = (k/m)(vT − v). ∫2vTv dv/(v − vT) = −(k/m)t → v = vT + vT e−kt/m. At t = (m/k) ln 2: v = vT(1 + 1/2) = 1.5vT.
A 0.20 kg ball is dropped from rest and experiences a resistive force F⃗r = −kv⃗ with k = 0.40 kg/s. Use g = 10 m/s². What is the ball's speed 0.50 s after it is released?
Answer and reasoning
A5.0 m/s A student who thinks the terminal speed is reached after one time constant, m/k = 0.50 s, picks this. At that time the speed is 1 − e⁻¹ ≈ 63 percent of vT. (Ignoring air resistance, gt, also gives 5.0 m/s.)
B1.8 m/s A student who uses the decaying form found for coasting objects, (mg/k)e−kt/m, picks this. That form fits an object slowing toward zero; a ball starting from rest has v = (mg/k)(1 − e−kt/m), which is zero at t = 0.
C3.2 m/sCorrect Separating variables in m dv/dt = mg − kv from v = 0 at t = 0 gives v = (mg/k)(1 − e−kt/m). Here mg/k = 5.0 m/s and m/k = 0.50 s, so v = (5.0 m/s)(1 − e⁻¹) = 3.2 m/s.
D1.1 m/s A student who uses k/m = 2.0 s as the time constant picks this: (5.0 m/s)(1 − e−0.25). The time constant is m/k = 0.50 s, which has units of seconds.
Working vT = mg/k = (0.20 kg)(10 m/s²)/(0.40 kg/s) = 5.0 m/s; m/k = 0.50 s. v = vT(1 − e−kt/m) = (5.0 m/s)(1 − e−1) = 3.2 m/s.
Two balls, A and B, of equal mass are dropped from rest. Each experiences a resistive force F⃗r = −kv⃗, with resistive constants kA and kB. The graph shows their speeds v as functions of time t. What is kA/kB?
Answer and reasoning
A2.00 A student who thinks a larger terminal speed means a larger resistive constant picks this: 8.0/4.0. At terminal speed the resistive force equals mg for both balls, so the faster ball has the smaller k.
B1.00 A student who reads the equal initial slopes as showing equal resistive constants picks this. At release v = 0, so the resistive force is zero and both start with acceleration g whatever their k.
C0.25 A student who uses the quadratic-drag model, in which the drag constant goes as 1/vT², picks this: (4.0/8.0)². For F⃗r = −kv⃗, k = mg/vT, which goes as 1/vT.
D0.50Correct At terminal speed kvT = mg, so for equal masses k is inversely proportional to vT. The graph shows vT = 8.0 m/s for A and 4.0 m/s for B, so kA/kB = 4.0/8.0 = 0.50.
Working k = mg/vT; equal m → kA/kB = vTB/vTA = (4.0 m/s)/(8.0 m/s) = 0.50.
A puck of mass m slides in a straight line across a level surface coated with a thin layer of oil. At t = 0 its speed is v₀, and the oil exerts a resistive force F⃗r = −kv⃗ on it; no other horizontal force is exerted on it. What is the magnitude of the puck’s acceleration at the instant it has traveled a distance mv₀/(2k)?
Answer and reasoning
A1.00 kv₀/m A student who treats the deceleration as constant at its initial value, kv₀/m, picks this. The resistive force, and so the deceleration, decreases as the puck slows: by the time the puck has traveled mv₀/(2k) its speed has halved, so its acceleration has halved too.
B0.50 kv₀/mCorrect Separating variables gives v = v₀e−kt/m, and integrating gives x = (mv₀/k)(1 − e−kt/m) = (m/k)(v₀ − v). At x = mv₀/(2k), v₀ − v = v₀/2, so v = v₀/2. The acceleration is dv/dt = −(k/m)v, so its magnitude at that instant is kv₀/(2m) = 0.50 kv₀/m.
C0.61 kv₀/m A student who finds the time by dividing the distance by the initial speed picks this: t = (mv₀/(2k))/v₀ = m/(2k), and then |a| = (kv₀/m)e−1/2 ≈ 0.61 kv₀/m. The puck moves more slowly than v₀ throughout, so it takes longer, (m/k) ln 2, to travel that distance.
D0.72 kv₀/m A student who divides the change in velocity by the elapsed time picks this: the speed falls by v₀/2 in (m/k) ln 2, giving kv₀/(2m ln 2) ≈ 0.72 kv₀/m. That is the average acceleration over the interval; the acceleration at the instant is dv/dt = −kv/m, which is smaller because the deceleration decreases as the puck slows.
Working Newton’s second law: m dv/dt = −kv. Separating variables and integrating from v₀ at t = 0: v = v₀e−kt/m. Integrating again from x = 0 at t = 0: x(t) = (mv₀/k)(1 − e−kt/m) = (m/k)(v₀ − v). With x = mv₀/(2k): v = v₀/2, reached at t = (m/k) ln 2. Then |a| = |dv/dt| = (k/m)v₀e−kt/m = kv/m = 0.50 kv₀/m (sympy-checked). Distractors: deceleration kept at its initial value, 1.00 kv₀/m; time found from x = v₀t, t = m/(2k), so |a| = (kv₀/m)e−1/2 ≈ 0.61 kv₀/m; average acceleration Δv/Δt = (v₀/2)/((m/k) ln 2) = kv₀/(2m ln 2) ≈ 0.72 kv₀/m.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
