The velocity of a material point at any instant differs from…
The velocity of a material point at any instant differs from the average velocity (from the beginning of motion to that instant) by a magnitude which is proportional to the kinetic energy of the point and inversely proportional to the time elapsed from the beginning. Find the dependence of distance on time.
Explanation of Method
Writing \(v=ds/dt\), \(\bar v = s/t\), and \(E_k=\tfrac12 mv^2\), the stated condition \(v-\bar v = k E_k/t\) becomes, after absorbing constants into \(\lambda=km/2\), the Clairaut-type equation \(s = tp-\lambda p^2\) with \(p=ds/dt\). Differentiating produces the usual Clairaut split: a general solution (a family of straight lines \(s=ct-\lambda c^2\)) and a singular solution (the envelope of that family, obtained by eliminating \(p\) between \(s=tp-\lambda p^2\) and \(t=2\lambda p\)). The physically meaningful branch is selected by the initial condition \(s(0)=0\): sympy's own solution of the ODE (\(\texttt{dsolve}\)) reproduces exactly the general-line family \(s=ct-\lambda c^2\), and forcing \(s(0)=0\) on that family forces \(c=0\) (pure rest, the trivial/degenerate branch). The genuine motion is therefore the singular solution, which was independently checked by direct substitution into \(t\,s'-s=\lambda(s')^2\): the residual simplifies to \(0\) identically.
1. Translate the physical statement into an ODE
With \(v=ds/dt=p\), \(\bar v=s/t\), \(E_k=\tfrac12mv^2\), and \(\lambda=km/2\), the condition \(v-\bar v=kE_k/t\) becomes \[ p-\frac{s}{t}=\frac{\lambda p^2}{t}\ \Longrightarrow\ tp-s=\lambda p^2\ \Longrightarrow\ s=tp-\lambda p^2, \] a Clairaut equation in \(s(t)\).
2. Differentiate and split into the two Clairaut branches
Differentiating \(s=tp-\lambda p^2\) with respect to \(t\) (using \(ds/dt=p\)) gives \(0=(t-2\lambda p)\,dp/dt\), so either \(dp/dt=0\) (general solution) or \(t=2\lambda p\) (singular solution). This was cross-checked by asking Sympy's \(\texttt{dsolve}\) to solve \(tp-s=\lambda p^2\) directly; it returns \(s(t)=\dfrac{t^2-(C_1+t)^2}{4\lambda}\), which expands to exactly the general-line family \(s=ct-\lambda c^2\) with \(c=-C_1/(2\lambda)\) — confirming this branch and showing Sympy's automatic solver does not surface the singular branch on its own, consistent with standard Clairaut-equation behaviour.
3. Case 1 (general solution): forced trivial by the initial condition
For \(dp/dt=0\), \(p=c\) is constant and \(s=ct-\lambda c^2\). Imposing \(s(0)=0\) gives \(-\lambda c^2=0\), i.e. \(c=0\), so this branch degenerates to \(s\equiv0\) — a point that never moves, which cannot be the intended physical motion.
4. Case 2 (singular solution): the genuine motion
The other factor gives \(t=2\lambda p\), i.e. \(p=t/(2\lambda)\). Substituting back into \(s=tp-\lambda p^2\): \[ s=t\cdot\frac{t}{2\lambda}-\lambda\left(\frac{t}{2\lambda}\right)^2=\frac{t^2}{2\lambda}-\frac{t^2}{4\lambda} =\frac{t^2}{4\lambda}. \] This automatically satisfies \(s(0)=0\).
5. Independent verification
Substituting \(s=t^2/(4\lambda)\) directly into the original (unsplit) equation \(t\,s'-s=\lambda (s')^2\) with Sympy: \(s'=t/(2\lambda)\), so \(t s'-s = t^2/(2\lambda)-t^2/(4\lambda)=t^2/(4\lambda)\) and \(\lambda(s')^2=\lambda\cdot t^2/(4\lambda^2) =t^2/(4\lambda)\) — the two sides agree identically (residual \(0\)), confirming the solution satisfies the ODE exactly, not just the split branches. The resulting velocity \(v=t/(2\lambda)\) is linear in \(t\), i.e. uniformly accelerated motion.
Result
Final answer:
With \(\lambda=km/2\) (\(k\) the proportionality constant, \(m\) the mass), the distance–time law is \[ \boxed{s(t)=\frac{t^2}{4\lambda}=\frac{t^2}{2km}}, \] i.e. uniformly accelerated motion with constant acceleration \(a=1/(2\lambda)\) and velocity \(v(t)=t/(2\lambda)\), starting from rest.