Solve a velocity-dependent resistive force
You will be able to: Separate variables in m dv/dt = −bv and use the initial velocity.
Why does resistance sometimes produce exponential slowing?
A cart starts at 6 m/s and experiences a horizontal resistive force proportional to its velocity. It loses a fixed fraction of its speed in each equal time interval, rather than losing the same number of metres per second.
A useful starting point: Springs in series and parallel →
Words and symbols before equations
- Linear resistance b
- Positive coefficient in F_r = −bv, with units kg/s or N·s/m.
- Differential equation
- An equation relating a function to its rate of change.
- Natural logarithm ln
- Inverse of the exponential function eˣ; d(ln v)/dv = 1/v for v > 0.
- Time constant τ
- Characteristic decay time m/b, measured in s.
What this picture assumes
Only horizontal linear resistance F = −bv; fixed mass, x(0) = 0 and positive initial velocity. Exact exponential solution; no finite-time stop or reversal in this ideal model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- τ = 2 s. At 2 s: v = 2.207 m/s; a = -1.104 m/s²; displacement 7.585 m. Limiting displacement 12 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For constant mass and no other horizontal force, m dv/dt = −bv. With positive initial velocity, rearrange to dv/v = −(b/m)dt. Integrate from v₀ to v and from 0 to t: ln(v/v₀) = −bt/m. Exponentiating gives v(t) = v₀e^(−t/τ). The zero-velocity solution is also valid and should not be lost by dividing by v.
Differentiate to obtain a = −(v₀/τ)e^(−t/τ). Integrate velocity with x(0) = x₀: x(t) = x₀ + v₀τ[1 − e^(−t/τ)]. Speed approaches zero and displacement approaches v₀τ without crossing either limit in this ideal model.
At one time constant, speed is v₀/e, about 36.8% of its initial value. A larger b shortens τ; a larger mass lengthens it. This prescribed linear resistance is not a universal law for air drag at all speeds.
A worked example, step by step
A 2 kg cart has b = 1 kg/s and v₀ = 6 m/s. Find v, a and displacement after 2 s.
- τ = m/b = 2 s, so t/τ = 1.
- v = 6/e ≈ 2.207 m/s.
- a = −v/τ ≈ −1.104 m/s².
- Δx = 6(2)(1 − 1/e) ≈ 7.585 m, below the limiting 12 m.
The acceleration is not constant: as speed falls, the resistive force magnitude falls too.
Does this ideal cart reverse direction because its acceleration is negative?
Compare with an explanation
No. Its positive speed approaches zero asymptotically; negative acceleration alone does not require reversal.
Predict. Change one thing. Explain.
Hold initial speed fixed. Change b while keeping mass fixed, then compare speed at one time constant in each case. Distinguish the fraction remaining from the elapsed time needed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
τ = 2 s. At 2 s: v = 2.207 m/s; a = -1.104 m/s²; displacement 7.585 m. Limiting displacement 12 m.
Only horizontal linear resistance F = −bv; fixed mass, x(0) = 0 and positive initial velocity. Exact exponential solution; no finite-time stop or reversal in this ideal model.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 3 kg object starts at 4 m/s with b = 1.5 kg/s and only linear resistance. Find τ, velocity at 2 s, limiting displacement and the limiting acceleration.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: τ = 3/1.5 = 2 s.
- 1 point: v(2) = 4/e ≈ 1.472 m/s.
- 1 point: Limiting displacement = v₀τ = 8 m.
- 1 point: Acceleration approaches zero from below as velocity approaches zero.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why does ln appear?
Separating variables requires integrating dv/v.
RECALL 2What sets the exponential time scale?
The ratio m/b.
RECALL 3Is linear drag assumed valid for all real fluids and speeds?
No. It is a specified model that must be justified for the situation.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Solve a velocity-dependent resistive force
- F_r = −bv; τ = m/b.
- v = v₀e^(−t/τ); a = −v/τ.
- x − x₀ = v₀τ[1 − e^(−t/τ)] with only linear resistance.
Remember: The acceleration is not constant: as speed falls, the resistive force magnitude falls too.
Conditions: Only horizontal linear resistance F = −bv; fixed mass, x(0) = 0 and positive initial velocity. Exact exponential solution; no finite-time stop or reversal in this ideal model.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.9.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.9, objectives 2.9.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about Solve a velocity-dependent resistive force. Your explanation and answers remain free to access.