Resistive Forces: Study Guide (Physics C: Mechanics 2.9)

A topic only in the calculus-based course. Physics 1 has no Topic 2.9 on resistive forces. Here the force depends on velocity, so Newton’s second law becomes a differential equation, and you need calculus to solve it. This topic follows Topic 2.8, Spring Forces, where the force depended on position instead.

What makes a force resistive

A resistive force depends on the object’s velocity and points opposite to the velocity. Air resistance on a falling seed, water drag on a coasting boat and the braking force from a magnet on a moving metal plate are all examples.

The model used in this course is a force proportional to velocity:

F_r = −kv

• k is a positive constant. It depends on the object’s size and shape and on the fluid (or magnet) it moves through.
• The minus sign makes the force point against v. If v is in +x, F_r is in −x. If v reverses, F_r reverses too.
• Units: k = F/v, so k is in N/(m/s) = N·s/m = kg/s.

Because F_r changes as v changes, the acceleration changes too. So the constant-acceleration equations from Topic 1.2 never apply to motion with a resistive force.

Newton’s second law becomes a differential equation

Write Newton’s second law as m dv/dt = ΣF. With a resistive force, v appears on both sides, so you get an equation for the function v(t), not a number.

Case A: only the resistive force acts along the motion. An object coasts on a level surface with no friction and no engine. Take +x along the initial velocity v₀:

m dv/dt = −kv

Case B: a constant force acts as well. An object falls from rest with drag. Take +y downward, so gravity is positive and drag is negative while the object moves down:

m dv/dt = mg − kv

Write the axis down first. Every sign in the equation depends on it.

Separation of variables

The method has four steps.

1. Put every v on one side and every t on the other.
2. Integrate both sides between matching limits: the lower limits are the initial state (t = 0, v = v₀) and the upper limits are a general state (t, v).
3. Use ∫ du/u = ln|u|.
4. Undo the logarithm with the exponential function.

Case A. dv/v = −(k/m) dt. Integrate from v₀ to v and from 0 to t:

ln(v/v₀) = −(k/m)t, so v(t) = v₀ e^(−t/τ), where τ = m/k.

τ is the time constant. It has units kg ÷ (kg/s) = s. After one time constant the velocity has fallen to e^(−1) ≈ 0.37 of v₀.

Case B. dv/(mg − kv) = dt/m. Let u = mg − kv, so du = −k dv. Integrating from 0 to v and from 0 to t gives −(1/k) ln[(mg − kv)/mg] = t/m. Rearranging:

v(t) = (mg/k)(1 − e^(−t/τ)), again with τ = m/k.

Acceleration and position from v(t)

Once you have v(t), use the calculus from Topic 1.2. Differentiate for acceleration. Integrate, with the initial position, for position.

	Case A: coasting, v₀ along +x	Case B: falling from rest, +y down
v(t)	v₀ e^(−t/τ)	v_T(1 − e^(−t/τ)), with v_T = mg/k
a(t) = dv/dt	−(v₀/τ) e^(−t/τ)	g e^(−t/τ)
position	x = x₀ + v₀τ(1 − e^(−t/τ))	y = y₀ + v_T t − v_Tτ(1 − e^(−t/τ))
as t → ∞	v → 0, a → 0, x → x₀ + v₀τ	v → v_T, a → 0, y keeps increasing at rate v_T

Each asymptote is set by the initial conditions and the forces. In Case A the object never quite stops, but its total displacement is finite: v₀τ = mv₀/k. In Case B the velocity heads towards v_T, whatever value it starts from.

Check a(t) in Case B at t = 0: a = g. At the instant of release v = 0, so there is no drag yet, and gravity alone accelerates the object.

Terminal velocity

When a constant force (such as gravity) and a resistive force act in opposite directions, the object speeds up until the resistive force grows to match the constant force. Then:

ΣF = 0, so mg − kv_T = 0, giving v_T = mg/k.

This is the terminal velocity: the greatest speed an object reaches under these forces when it starts from rest (or from any speed below v_T). The acceleration is then zero, so the velocity stops changing.

If an object starts faster than v_T (thrown down hard, say), the resistive force is bigger than mg. The net force points up, against the motion, and the object slows down towards v_T. The same equation, solved with v(0) = v₀, gives:

v(t) = v_T + (v₀ − v_T) e^(−t/τ)

Either way, the asymptote is v_T. The time constant also has a graph meaning: the tangent to the v–t curve at t = 0 meets the asymptote at t = τ. For falling from rest, τ = v_T/g.

Worked example 1: a coasting underwater drone

Question. Take +x along the drone’s initial velocity, origin where the motor stops. A 6.0 kg underwater drone moves horizontally at 1.2 m/s when its motor cuts out. Model the water drag as F_r = −kv with k = 3.0 kg/s, and ignore all other horizontal forces. Find (a) v(t), (b) the initial acceleration, (c) the time for the speed to halve, (d) the velocity and position at t = 3.0 s and (e) the greatest distance the drone can coast.

1. Equation. m dv/dt = −kv. Separate: dv/v = −(k/m) dt.
2. (a) Integrate from (0, 1.2 m/s) to (t, v): ln(v/1.2) = −t/τ with τ = m/k = 6.0/3.0 = 2.0 s. So v = 1.2 e^(−t/2.0) (m/s, t in s).
3. (b) a = dv/dt = −(1.2/2.0) e^(−t/2.0). At t = 0: a = −0.60 m/s². Check with forces: −kv₀/m = −3.0 × 1.2 ÷ 6.0 = −0.60 m/s².
4. (c) v = v₀/2 when e^(−t/τ) = ½, so t = τ ln 2 = 2.0 × 0.693 = 1.4 s.
5. (d) At 3.0 s: v = 1.2 e^(−1.5) = 0.27 m/s. Position: x = v₀τ(1 − e^(−t/τ)) = 1.2 × 2.0 × (1 − e^(−1.5)) = 1.9 m. Acceleration then is −0.13 m/s², smaller because the drone is slower.
6. (e) As t → ∞, x → v₀τ = mv₀/k = 2.4 m.

Interpretation. The drone never quite stops, but it never passes 2.4 m. A constant-acceleration answer, using a = −0.60 m/s² throughout, would say it stops after 2.0 s and 1.2 m. That is wrong: the drag weakens as the drone slows, so it coasts further and for longer.

Worked example 2: a falling seed pod

Question. Take +y downward, origin at the release point. A 0.020 kg seed pod is released from rest. Model the air resistance as F_r = −kv with k = 0.040 kg/s. Find (a) the terminal velocity and the time constant, (b) v(t), (c) the velocity and distance fallen after 1.0 s, compared with free fall, and (d) the acceleration at t = 0 and at t = 1.0 s.

1. (a) At terminal velocity ΣF = 0: mg = kv_T, so v_T = mg/k = 0.020 × 9.8 ÷ 0.040 = 4.9 m/s. τ = m/k = 0.020 ÷ 0.040 = 0.50 s.
2. (b) Newton’s second law: m dv/dt = mg − kv. Separate and integrate from (0, 0) to (t, v), as in Case B: v = 4.9(1 − e^(−t/0.50)) (m/s, t in s).
3. (c) At 1.0 s: v = 4.9(1 − e^(−2.0)) = 4.2 m/s. Integrate for the distance: y = v_T t − v_Tτ(1 − e^(−t/τ)) = 4.9 − 4.9 × 0.50 × (1 − e^(−2.0)) = 2.8 m. In free fall it would have reached 9.8 m/s and fallen 4.9 m.
4. (d) a = dv/dt = g e^(−t/τ). At t = 0, a = 9.8 m/s²: no drag yet. At 1.0 s, a = 9.8 e^(−2.0) = 1.3 m/s².

Check with forces at 1.0 s. Drag = kv = 0.040 × 4.24 = 0.170 N; weight = 0.196 N; net force = 0.027 N down, and 0.027 N ÷ 0.020 kg ≈ 1.3 m/s². The calculus and the free-body diagram agree.

How long to “reach” v_T? Never exactly. It reaches 90% of v_T when e^(−t/τ) = 0.10, at t = τ ln 10 ≈ 1.2 s, and 99% at about 2.3 s.

Common misconceptions

• “At terminal velocity there is no force on the object.” Gravity and drag both still act. They are equal and opposite, so the net force is zero.
• “Drag always points up.” It points opposite to the velocity. On a ball thrown upwards, drag points down, adding to gravity, until the top.
• “The acceleration is zero at the start because the object is not moving yet.” Released from rest, a = g at t = 0. The acceleration falls to zero as v approaches v_T.
• “A heavier object of the same shape has a bigger acceleration at release.” Both start with a = g. The heavier one has the larger v_T = mg/k and takes longer (τ = m/k) to approach it.
• Using the constant-acceleration equations. Acceleration changes with v, so integrate (Worked example 1).
• Mismatched limits. If the lower limit of the v integral is v₀, the lower limit of the t integral must be the time when v = v₀.
• “Infinite time means infinite distance.” A coasting object never stops, yet its displacement approaches the finite value mv₀/k.

Where this leads

Next, Topic 2.10, Circular Motion, applies Newton’s second law along a curved path. The same separation-of-variables method returns whenever a rate of change depends on the quantity itself. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.