I heard that the formula- $v_{\text{Final}}^2 = v_{\text{Initial}}^2 + 2ad$

(Velocity final squared = velocity initial squared plus two times acceleration and displacement)

$t$ = time

is a combined formula of these two:

$v_{\text{Final}} = v_{\text{Initial}} + at$

$d = v_{\text{Initial}}t + \frac12 at^2$

if true, how so?

  • $\begingroup$ Yes, it is correct. Express $t$ from the first of the two equations, and plug it into the second, in both places. Then simplify. You might benefit from noticing $V_f^2-V_0^2=(V_f-V_0)(V_f+V_0)$. You might also benefit from learning how to write equations here (using what is essentially LaTeX syntax). See, e.g., this short guide: <sharelatex.com/learn/Mathematical_expressions> or this more detailed guide: <ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf> $\endgroup$ – StR Oct 28 '16 at 1:44

The first formula is essentially the work-energy theorem, slightly reworked.

If a net force $F$ does work $W$ on a massive object, its kinetic energy $K$ increases:

$$W=\Delta K$$

If the force is constant it causes uniform acceleration $a$ and the work done is:

$$W=ma\Delta x\tag{1}$$

Where $\Delta x$ is the displacement. The increase in kinetic energy is:

$$\Delta K=K_f-K_0=\frac12 mv_f^2-\frac12 mv_0^2\tag{2}$$

With $(1)=(2)$:

$$ma\Delta x=\frac12 mv_f^2-\frac12 mv_0^2$$

Divide both sides by $m$ and multiply by $2$:

$$v_f^2=v_0^2+2a\Delta x\tag{3}$$

By eliminating $t$ from your second and third formulas, you get $(3)$.


yes it's true

$$v_{Final}=v_{Initial}+at$$ $$\Rightarrow t=\dfrac{v_{Final}- v_{Initial}}{a}........(1)$$ $$d=v_{Initial}t+\dfrac{1}{2}at^2.........(2)$$ Substitute vale of $t$ from equation(1) in equation(2) $$d=v_{Initial}\bigg(\dfrac{v_{Final}- v_{Initial}}{a}\bigg)+\dfrac{1}{2}a\Bigg(\dfrac{v_{Final}- v_{Initial}}{a}\bigg)^2$$ $$v_{Final}=v_f\;\;\;\;\;\;\;\;\;\;\;v_{Initial}=v_i$$ $$d=v_i\bigg(\dfrac{v_f- v_i}{a}\bigg)+\dfrac{1}{2}a\Bigg(\dfrac{v_f- v_i}{a}\bigg)^2$$ $$d=\dfrac{(v_fv_i-v_i^2)}{a}+\dfrac{1\not a}{2\not a^2}(v_f^2+v_i^2-2v_fv_i)$$ $$2ad=2v_fv_i-2v_i^2+v_f^2+v_i^2-2v_fv_i$$ $$2ad=v_f^2-v_i^2$$ $$v_f^2=v_i^2+2ad$$ $$v_{Final}^2=v_{Initial}^2+2ad$$


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