Resistive forces

A resistive force opposes motion through a fluid like air or water. Unlike a steady push, its size depends on how fast the object is moving: the faster it goes, the harder the fluid pushes back. The direction is always opposite the velocity, so for an object falling straight down, the drag points straight up.

The simplest useful model is linear drag, $F_d = bv$, where $v$ is the speed and $b$ is a constant set by the object and the fluid. Double the speed and you double the drag. (For faster motion through air a quadratic model $F_d = cv^2$ fits better; the ideas are the same, and we will use linear drag for the calculations to keep the algebra clean.)

This is what sets drag apart from the friction of Topic 2.7. Kinetic friction has a fixed size $\mu_k N$ that does not care how fast the block slides. Drag has no fixed size at all; it climbs as the object speeds up. That one difference, a force that grows with speed, is the source of every result and every mistake in this topic.

Two questions run through the topic: how does the speed change over time, and what speed does the object end up at? Both come from a single equation of motion (§3), and the answer to the second is the terminal velocity (§4).

The most common drag mistake is to treat it like a constant force and reach for constant-acceleration kinematics, $v = v_0 + at$ and the rest. Those formulas need a fixed acceleration, and a falling object with drag does not have one.

Watch the acceleration over the fall. At the instant of release the object is at rest, so $v = 0$, the drag $bv$ is zero, and the only force is gravity. The acceleration starts at the full value $a = g$. As the object speeds up, the drag grows, the net force $mg - bv$ shrinks, and the acceleration drops, heading toward zero as the speed levels off.

So the acceleration is largest at the start and smallest at the end, the opposite of what people often guess, and it is never constant. A velocity-versus-time graph is not a straight line; it is a curve that rises steeply at first and then bends over and flattens. Any answer that uses $v = gt$ or a fixed $a$ is the constant-force trap.

The drag itself rises with speed, never the reverse. It is tempting to say the object slows its gain because the drag is easing off, but the gain eases because the drag is growing, not shrinking.

Every result in this topic comes from one equation of motion, so it is worth writing carefully. Pick a positive direction and stick with it. Taking downward as positive for a falling object, gravity acts in the $+$ direction with size $mg$, and drag opposes the motion, so it acts in the $-$ direction with size $bv$. Newton's second law is then

$m\dfrac{dv}{dt} = mg - bv.$

The two signs are the whole game. Gravity is $+mg$ because it drives the fall; drag is $-bv$ because it fights the fall. Flip the drag to $+bv$ and the equation claims the fluid speeds the object up, which is nonsense and gives a negative or runaway terminal velocity. Flip gravity, or write $bv - mg$, and the fall runs backward.

A quick check guards the signs: at $v = 0$ the equation must give $m\,dv/dt = mg$, a positive (downward) acceleration of $g$. If your equation gives anything else at release, a sign is wrong. Get this line right and the terminal velocity (§4) and the full solution (§5) follow directly.

As the object speeds up, the drag grows until it balances gravity. At that point the net force is zero, the acceleration is zero, and the speed stops changing. That steady speed is the terminal velocity $v_t$. The College Board states the skill directly: describe the motion of an object acted upon by a resistive force, and terminal velocity is the heart of it.

Find it by setting the acceleration to zero in the equation of motion. With $dv/dt = 0$, $mg - bv_t = 0$, so

$v_t = \dfrac{mg}{b}.$

Here is the idea that costs the most points. Terminal velocity is zero net force, not zero force. Gravity has not switched off; it still pulls down with $mg$. The drag has simply grown until it pushes up with the same $mg$. Two real forces, equal and opposite, cancel to a zero net force, which is why the speed holds steady. Saying no force acts at terminal velocity, or that gravity is gone, describes a different and impossible situation.

For the quadratic model $F_d = cv^2$ the same balance gives $cv_t^2 = mg$, so $v_t = \sqrt{mg/c}$. The number is different and the route has a square root, but the physics is identical: terminal velocity is where the drag has grown to match gravity and the net force is zero.

The equation of motion $m\,dv/dt = mg - bv$ gives the full speed-versus-time curve, not just the final value. Because the right side depends on $v$, separate the variables and integrate. Writing $v_t = mg/b$ and $\tau = m/b$, the equation becomes $dv/dt = (v_t - v)/\tau$, which separates to

$\displaystyle \int \dfrac{dv}{v_t - v} = \int \dfrac{dt}{\tau}.$

Integrating and applying $v = 0$ at $t = 0$ gives the standard result

$v(t) = v_t\left(1 - e^{-t/\tau}\right), \qquad \tau = \dfrac{m}{b}.$

Read the curve. At $t = 0$ it gives $v = 0$, the release from rest. For large $t$ the exponential dies and $v \to v_t$, the terminal velocity, approached but never quite reached. The constant $\tau = m/b$ is the time constant: after one $\tau$ the object has gained about $63\%$ of $v_t$ (since $1 - e^{-1} \approx 0.63$), about $86\%$ after $2\tau$, and about $95\%$ after $3\tau$. A heavier object, or a smaller drag constant, means a larger $\tau$ and a slower approach.

Setup. A $2$ kg ball is dropped from rest and falls with linear drag, $b = 0.5$ N$\cdot$s/m. Take $g = 10$ m/s$^2$. (a) Find the terminal velocity and the time constant. (b) Find the speed and acceleration at $t = \tau$. (c) Check what gravity and drag are doing at terminal velocity.

(a) Terminal velocity and time constant. Set the net force to zero: $v_t = mg/b = (2 \cdot 10)/0.5 = 40$ m/s. The time constant is $\tau = m/b = 2/0.5 = 4$ s.

(b) Speed and acceleration at $t = \tau$. Using $v(t) = v_t(1 - e^{-t/\tau})$ at $t = \tau$: $v = 40\,(1 - e^{-1}) \approx 40 \cdot 0.63 \approx 25$ m/s. The acceleration there is $a = (mg - bv)/m = (20 - 0.5 \cdot 25)/2 \approx 3.75$ m/s$^2$, already well below the $10$ m/s$^2$ it started at.

(c) At terminal velocity. Gravity is still $mg = 20$ N, unchanged from the moment of release. The drag has grown to $bv_t = 0.5 \cdot 40 = 20$ N, exactly matching it. The net force is $20 - 20 = 0$, so the acceleration is zero and the speed holds at $40$ m/s. Nothing turned gravity off; the drag simply caught up to it.

The trap to avoid. At terminal velocity the net force is zero, yet the weight is still $20$ N and the drag is still $20$ N. Reading zero net force as no force loses the whole picture, and reaching for $v = gt$ at any point treats a changing acceleration as if it were fixed.