Proof of the time-independent equation

Problem

Starting from the laws of uniformly accelerated motion $s(t) = s_0 + v_0 t + \tfrac{1}{2} a t^2$ and $v(t) = v_0 + a t$, prove the time-independent relation $v^2 = v_0^2 + 2a\,\Delta s$. Show the steps explicitly.

Solution

Strategy. The idea is to eliminate the time $t$ between the two laws of motion, so as to obtain a relation that directly links velocity and displacement.

Step 1 — Obtain the time from the velocity law. From the relation $v = v_0 + a t$, isolating $t$ (assuming $a \neq 0$):

$t = \frac{v - v_0}{a}$

Step 2 — Express the displacement. Let $\Delta s = s - s_0$ denote the displacement; from the equation of motion:

$\Delta s = v_0 t + \tfrac{1}{2} a t^2$

Step 3 — Substitute the time. We insert the expression for $t$ found in step 1:

$\Delta s = v_0 \cdot \frac{v - v_0}{a} + \tfrac{1}{2} a \left(\frac{v - v_0}{a}\right)^2 = \frac{v_0(v - v_0)}{a} + \frac{(v - v_0)^2}{2a}$

Step 4 — Expand and simplify. Reducing to the common denominator $2a$ and expanding the squares:

$\Delta s = \frac{2v_0 v - 2v_0^2 + v^2 - 2 v_0 v + v_0^2}{2a} = \frac{v^2 - v_0^2}{2a}$

Step 5 — Isolate $v^2$. Multiplying both sides by $2a$:

$2a\,\Delta s = v^2 - v_0^2$

from which we obtain the sought relation, independent of time:

$\ev{v^2 = v_0^2 + 2a\,\Delta s}$

This formula is particularly useful when the time is neither known nor required, for example when calculating stopping distances or maximum heights.

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato Skills: Impostazione simbolica