GRE Physics Companion: Drag Forces and Terminal Velocity

Drag problems are usually tests of force direction, terminal-state reasoning, and proportional scaling. The quickest reliable method is to identify the velocity relative to the fluid, choose the stated drag law, and then decide whether the problem concerns instantaneous acceleration or the terminal condition.

1 Core relations

Relative velocity is

|--------------------| |vrel = vbody − vfluid. ---------------------

(1)

Linear drag is

|------------| |Fd = − bvrel.| --------------

(2)

Quadratic drag is

|-----------------| Fd = − c|vrel|vrel. | ------------------

(3)

For downward terminal motion with negligible buoyancy,

|----------| |F = mg. | --d--------

(4)

Thus

|--------| | mg | vt = --b-| ----------

(5)

for linear drag and

|----∘-----| | mg-| vt = c | ------------

(6)

for quadratic drag.

Figure 1. A compact drag-problem workflow: form the fluid-relative velocity, identify the drag law, determine direction, then use Newton’s second law or the terminal-force balance.

2 High-value quadratic scaling

When

1- 2 Fd = 2 ρCDAv ,

(7)

terminal speed is

|--------------| | ∘ -------| |vt = -2mg--.| --------ρCDA----

(8)

Therefore

v ∝ √m,- v ∝ A−1∕2, v ∝ ρ− 1∕2, v ∝ C − 1∕2. t t t t D

(9)

Figure 2. Quadratic-drag terminal-speed scaling. Multiplicative changes in mass or drag parameters enter through square roots.

3 Worked GRE example 1: scaling without recomputing

A falling object is modeled with quadratic drag. Its frontal area is increased by a factor of 9, while mass, fluid density, and C_D remain unchanged. How does its terminal speed change?

Since

vt ∝ √1--, A

(10)

we have

vt,2-= √1--= 1-. vt,1 9 3

(11)

Therefore

|------------| | 1 | |vt,2 = --vt,1.| -------3-----

(12)

4 Worked GRE example 2: instantaneous acceleration below terminal speed

A body falls downward under quadratic drag and has terminal speed 30.0 m∕s. Find its downward acceleration when its speed is 18.0 m∕s.

For quadratic drag,

( v2) a = g 1 − -2- . vt

(13)

Therefore

[ ( ) ] 18.0 2 2 a = 9.81 1 − 30.0 = 6.28 m ∕s.

(14)

Thus

|--------------------------| |a = 6.28 m ∕s2 downward. | ---------------------------

(15)

5 GRE speed questions

A falling body has reached terminal velocity. Which statement is correct? (A) Its velocity is zero. (B) Its acceleration is zero. (C) The drag force is zero. (D) Gravity no longer acts.
Under quadratic drag, the mass of a falling body is multiplied by 4 while all drag parameters remain unchanged. Its terminal speed becomes (A) half as large (B) unchanged (C) twice as large (D) four times as large.
Under quadratic drag, the product C_DA is multiplied by 16. The terminal speed becomes (A) one-sixteenth as large (B) one-fourth as large (C) half as large (D) four times as large.
A runner moves east at 8 m∕s while the air moves east at 3 m∕s. The runner’s air-relative speed is (A) 3 m∕s (B) 5 m∕s (C) 8 m∕s (D) 11 m∕s.

6 Answers and rationales

B. Terminal velocity is a zero-net-force state, so acceleration is zero while velocity remains nonzero.
C. Quadratic-drag terminal speed scales as .
B. Quadratic-drag terminal speed scales as (C_DA)^−1∕2.
B. The relative velocity is 8 − 3 = 5 m∕s east.

7 Common GRE traps

Setting velocity to zero at terminal speed instead of acceleration.
Using the object’s ground speed instead of its fluid-relative speed.
Forgetting that quadratic-drag terminal speed contains a square root.
Using F_d = mg away from terminal conditions.
Assuming drag always points upward; it opposes relative motion, whatever that direction is.

References

[1] PhysicsLibrary, M02-10, Drag Forces and Terminal Velocity.

[2] PhysicsLibrary, M02-01, Newton’s Laws of Motion.

[3] OpenStax, University Physics, Volume 1, CC BY 4.0.

"GRE Physics Companion: Drag Forces and Terminal Velocity" is owned by bloftin.
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