Question

A particle of mass m is acted upon by a force F given by the empirical law $F = \dfrac{R}{{{t^2}}}v(t)$. If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot:
A. $\log {\text{ }}v(t){\text{ against }}\dfrac{1}{t}$
B. $v(t){\text{ against }}{{\text{t}}^2}$
C. $\log {\text{ }}v(t){\text{ against }}\dfrac{1}{{{t^2}}}$
D. $\log {\text{ }}v(t){\text{ against t}}$

Answer 1

Hint: If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity. The rate of change of velocity gives us acceleration. Force will be the product of mass and acceleration.
Formula used:
$\eqalign{
  & F = ma \cr
  & a = \dfrac{{dv}}{{dt}} \cr} $

Complete step-by-step solution:
We were given the force equation and that is dependent on the velocity at every instant. We can write the force as the product of mass and acceleration and acceleration as the rate of a range of velocity.
After that we will integrate the given expression to get the velocity in terms of the time and some constant. We have force expression as
$F = \dfrac{R}{{{t^2}}}v(t)$
We can express this in terms of mass and acceleration as
$\eqalign{
  & F = ma \cr
  & \Rightarrow a = \dfrac{{dv}}{{dt}} \cr} $
By using the above information and by substituting that in the equation $F = \dfrac{R}{{{t^2}}}v(t)$ we will get
$\eqalign{
  & F = \dfrac{R}{{{t^2}}}v(t) \cr
  & \Rightarrow m\dfrac{{dv}}{{dt}} = \dfrac{R}{{{t^2}}}v(t) \cr
  & \Rightarrow \smallint \dfrac{{dv}}{v} = \smallint \dfrac{{Rdt}}{{m{t^2}}} \cr
  & \Rightarrow ln\;v = \dfrac{{ - R}}{{mt}} \cr
  & \Rightarrow ln\;v\;\; \propto \;\; \dfrac{1}{t} \cr} $
From the final result we had got R, m are the constants and they don’t vary with any other terms. So we can conclude that the logarithm of velocity will be inversely proportional to the time. Hence if we plot the logarithm of the velocity graph against the inverse of time, it would be convenient for us to determine the characteristics of motion of the particle.
Hence option A will be the answer.


Note: If the product of two variables is constant then the plot between them will be a hyperbola. So according to the final solution we had got, since the logarithm of velocity is inversely proportional to time, the product of that logarithm of velocity function and the time would be the hyperbola.