A particle of mass $$m$$ is acted upon by a force $$F$$ given by the empirical law $$F = \frac{R}{t^2}v(t)$$. If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot

We are given the force $$ F = \frac{R}{t^2} v(t) $$ acting on a particle of mass $$ m $$. According to Newton's second law, force is also equal to mass times acceleration, so $$ F = m \frac{dv}{dt} $$. Setting these equal, we get:

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

To solve this differential equation, we separate the variables. Divide both sides by $$ v $$ (assuming $$ v \neq 0 $$) and multiply by $$ dt $$:

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

Now, integrate both sides. The left side with respect to $$ v $$ and the right side with respect to $$ t $$:

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

Integrating the left side: $$ m \int \frac{1}{v} dv = m \ln|v| $$. Integrating the right side: $$ R \int t^{-2} dt = R \left( \frac{t^{-1}}{-1} \right) = -\frac{R}{t} $$. So we have:

$$ m \ln|v| = -\frac{R}{t} + C $$

where $$ C $$ is the constant of integration. Since the particle starts from rest, as $$ t \to 0^+ $$, $$ v \to 0 $$. As $$ t \to 0^+ $$, $$ -\frac{R}{t} \to -\infty $$, so $$ \ln|v| \to -\infty $$, which implies $$ v \to 0 $$, satisfying the initial condition.

For $$ t > 0 $$ and $$ v > 0 $$, we can drop the absolute value:

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

Solve for $$ \ln v $$ by dividing both sides by $$ m $$:

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

Let $$ K = \frac{C}{m} $$, so:

$$ \ln v = -\frac{R}{m t} + K $$

Exponentiate both sides to solve for $$ v $$:

$$ v = e^K e^{-\frac{R}{m t}} $$

Let $$ A = e^K $$, so:

$$ v = A e^{-\frac{R}{m t}} $$

To test this law experimentally, we need a linear relationship that can be verified with a straight-line plot. Take the natural logarithm of both sides:

$$ \ln v = \ln A - \frac{R}{m t} $$

This equation is of the form $$ y = mx + b $$, where $$ y = \ln v $$ and $$ x = \frac{1}{t} $$. The slope is $$ -\frac{R}{m} $$ and the intercept is $$ \ln A $$. Thus, plotting $$ \ln v $$ against $$ \frac{1}{t} $$ should yield a straight line.

In experimental physics, "log" often means base 10 logarithm. Using $$ \log_{10} v $$:

$$ \log_{10} v = \frac{\ln v}{\ln 10} = \frac{1}{\ln 10} \left( \ln A - \frac{R}{m t} \right) = -\frac{R}{m \ln 10} \cdot \frac{1}{t} + \frac{\ln A}{\ln 10} $$

This is still linear in $$ \frac{1}{t} $$, with slope $$ -\frac{R}{m \ln 10} $$ and intercept $$ \frac{\ln A}{\ln 10} $$. Therefore, plotting $$ \log v(t) $$ against $$ \frac{1}{t} $$ will give a straight line.

Now, evaluate the options:

• Option A: $$ \log v(t) $$ against $$ \frac{1}{t} $$ - This matches our derived linear relationship.
• Option B: $$ v(t) $$ against $$ t^2 $$ - From $$ v = A e^{-\frac{R}{m t}} $$, this is not linear in $$ t^2 $$.
• Option C: $$ \log v(t) $$ against $$ \frac{1}{t^2} $$ - Our equation has $$ \log v $$ linear in $$ \frac{1}{t} $$, not $$ \frac{1}{t^2} $$. Plotting against $$ \frac{1}{t^2} $$ would not yield a straight line.
• Option D: $$ \log v(t) $$ against $$ t $$ - Our equation shows $$ \log v $$ depends on $$ \frac{1}{t} $$, not directly on $$ t $$. This would not be linear.

Hence, the best way to test the law experimentally is to plot $$ \log v(t) $$ against $$ \frac{1}{t} $$, which is Option A.

Hence, the correct answer is Option A.