1. Physical situation

A flat plate moves normal to its plane through a very dilute gas.

• Gas pressure is very low $\Rightarrow$ collisions among molecules can be neglected compared to molecule-plate collisions.
• Plate speed $v$ is much smaller than average molecular speed $u$, i.e. $v \ll u$.
• A constant external force $F$ pulls the plate.

We must determine how the gas exerts a resistive force on the plate.

2. Pressure on leading and trailing faces

Because the plate is moving through the gas:

• On the leading face, molecules strike the plate more frequently and with slightly larger relative speed.
• On the trailing face, molecules catch up less effectively, so pressure is slightly smaller.

For $v \ll u$, the change in pressure can be found from kinetic theory.

Pressure due to molecular impacts scales like $p \sim \rho (\text{relative speed})^2$ more carefully, the difference in momentum transfer rate on the two sides is linear in $v$ for small $v$.

Since molecular thermal speed is $u$, the first-order correction to pressure is of the form $\Delta p \propto \rho u v.$ Thus, $\Delta p \propto uv$ (up to constants and density factors).

So Option A is true.

3. Resistive force on the plate

If the area of the plate is $A$, then resistive force due to pressure difference is $F_r = A\,\Delta p.$ Since $\Delta p \propto uv,$ and for the gas $u$ is essentially constant, $F_r \propto v.$

Therefore the drag force is proportional to speed for $v \ll u$.

So Option B is true.

4. Equation of motion of the plate

The plate is pulled by constant force $F$, while gas exerts resistive force proportional to $v$: $F_r = kv$ for some constant $k$.

Hence, $m\frac{dv}{dt} = F-kv.$ So acceleration is $a = \frac{dv}{dt} = \frac{F-kv}{m}.$

This is not constant because it depends on $v$.

Initially, if $v$ is small, acceleration is nearly $F/m$. As $v$ increases, resistive force increases, and acceleration decreases.

Therefore the plate does not continue with constant non-zero acceleration at all times.

So Option C is false.

5. Long-time behavior

At later times, as $v$ increases, the resistive force $kv$ increases until it equals the applied force $F$.

At that stage, $F = kv,$ so $a=0.$ The plate then moves with terminal speed $v_t = \frac{F}{k}.$

Thus at a later time the external force balances the resistive force.

So Option D is true.

6. Evaluation of all options

• A: True
• B: True
• C: False
• D: True

Hence the correct options are $\boxed{A,\ B,\ D}$

7. Comparison with stored answer

Stored correct answer: A, B, D

My derived answer: A, B, D

They match.