1. Write Newton's second law

Given empirical force law: $F=\frac{R}{t^2}v(t)$

Using Newton's second law, $m\frac{dv}{dt}=\frac{R}{t^2}v$

So the differential equation is $\frac{dv}{dt}=\frac{R}{m}\frac{v}{t^2}$

2. Separate variables

$\frac{1}{v}dv=\frac{R}{m}\frac{1}{t^2}dt$

Integrating, $\int \frac{1}{v}dv=\frac{R}{m}\int t^{-2}dt$

$\ln v=\frac{R}{m}(-t^{-1})+C$

Hence, $\ln v = -\frac{R}{m}\frac{1}{t}+C$

or $\log v = -\left(\frac{R}{m}\cdot \frac{1}{2.303}\right)\frac{1}{t}+C'$

3. Interpret the result

This is of the form $\log v = \text{constant} + \text{(constant)}\left(\frac{1}{t}\right)$

Therefore, a plot of $\log v(t)$ against $\dfrac{1}{t}$ will be a straight line.

4. Check options

• A: $v(t)$ against $t^2$ → not linear
• B: $\log v(t)$ against $\dfrac{1}{t^2}$ → not linear from the derived equation
• C: $\log v(t)$ against $t$ → not linear
• D: $\log v(t)$ against $\dfrac{1}{t}$ → linear

So the best plot is Option D.

5. About “starting from rest”

The mathematical form gives $v=Ae^{-R/(mt)}$ which tends to $0$ as $t\to 0^+$, so motion can be consistent with starting from rest in the limiting sense.

Therefore the experimentally testable linear relation is still with $\log v$ vs $1/t$.

Final Answer: $\boxed{\text{D}}$