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Step-by-Step Solution
Step 1: Write down the given force law
The empirical law states that the force on the particle is given by
$F = \frac{R}{t^2} \, v(t)$,
where $R$ is a constant, $t$ is the time, and $v(t)$ is the velocity at time $t$.
Step 2: Express the force using Newton’s second law
Newton’s second law tells us that
$F = m \frac{dv}{dt}$.
Therefore,
$m \frac{dv}{dt} = \frac{R}{t^2} \, v(t).$
Step 3: Rearrange and separate the variables
Divide both sides by $v(t)$ and multiply by $dt$ to get
$\displaystyle \frac{dv}{v} = \frac{R}{m} \frac{dt}{t^2}.$
Step 4: Integrate both sides
Integrate:
$\displaystyle \int \frac{dv}{v} = \frac{R}{m} \int \frac{dt}{t^2}.$
On the left side,
$\int \frac{dv}{v} = \ln v.$
On the right side,
$\int \frac{dt}{t^2} = -\frac{1}{t}.$
So we get
$\ln v = \frac{R}{m} \left( -\frac{1}{t} \right) + C,$
where $C$ is the constant of integration.
Step 5: Rewrite the expression for $v$
The above equation simplifies to
$\ln v = -\, \frac{R}{m} \,\frac{1}{t} + C.$
We can write this in the form
$\ln v = -\,\frac{R}{m} \left(\frac{1}{t}\right) + C.$
Step 6: Interpret the form of the relationship
A straight line can be written in the form $y = mx + b.$ Here, if we set
$y = \ln v$ and $x = \frac{1}{t},$
then the equation becomes
$\ln v = -\,\frac{R}{m} \left(\frac{1}{t}\right) + C,$
a straight-line equation, with slope $-\frac{R}{m}$ and intercept $C.$
Step 7: Conclude the best experimental plot
The result shows that a plot of $\ln v(t)$ (on the vertical axis) against $\frac{1}{t}$ (on the horizontal axis) should be a straight line if the law
$F = \frac{R}{t^2} \, v(t)$
is correct. Thus, plotting
$\ln v(t)$
against
$\frac{1}{t}$
is the best way to test the given law experimentally.
