I'm reading this book on Classical mechanics, and they show a derivation of the mentioned kinematic formula: $V_f^2=V_i^2+2a\left(x_f-x_0\right)$

They start by saying that the acceleration of an object down an incline is: $$\ddot{x}=gsin\theta$$, which I completelly understand. Then they mention:

"We can find the velocity of the object after it moves from rest a distnce $\left(x_f-x_0\right)$ down the incline by multiplying $\ddot{x}=gsin\theta$ by $2\dot{x}$ and integrating"

$$2\dot{x}\ddot{x}=2\dot{x}gsin\theta\tag{1}$$ $$\frac{d}{dt}\dot{x}^2=2gsin\theta\frac{dx}{dt}\tag{2}$$ $$\int_{0}^{v_f^2}d\left(\dot{x}^2\right)=2gsin\theta\int_{0}^{x_f-x_0}dx\tag{3}$$

At $t=0$,both $x_0=\dot{x}=0$, and, at $t=t_{final}$,$x=x_f-x_0$ and the velocity $\dot{x}=v_f$ $$v_f^2=2gsin\theta\left(x_f-x_0\right)$$

My questions are:

1) How they start with or assume equation $\left(1\right)$ ?

2) How they go from $2\dot{x}\ddot{x}$ to $\frac{d}{dt}\dot{x}^2$ ?

  • $\begingroup$ You quoted the sentence where they tell you how they obtain $(1)$! For 2., just use the chain rule. $\endgroup$ – ACuriousMind Jun 23 '15 at 22:24

Equation (1) is just $\ddot x=g\sin\theta$ multiplied by $2\dot x$ on both sides. If you understand $\ddot x=g\sin\theta$, this should follow directly. For your second question, we use the chain rule, $$\frac{d}{dt}\dot x^2=\frac{d\dot x^2}{d\dot x}\frac{d\dot x}{dt}=2\dot x\ddot x.$$

  • $\begingroup$ Whoops, that's what I meant. $\endgroup$ – Alex S Jun 23 '15 at 22:52
  • $\begingroup$ Thanks Alex, but my question on equation #1 was actually referring to how they assumed multiplying $\ddot{x}=gsin\theta$ by $2\dot{x}$ was going to lead to something $\endgroup$ – Hermes Chirino Jun 23 '15 at 22:57
  • $\begingroup$ @HermesChirino A lot of physics is about knowing how to guess the right thing, and recognizing patterns. I expect that wherever the authors of your textbook saw this derivation (or if they did it themselves), they recognized that the expression they wanted had an $x$ on the right, so they needed to get one there. They also needed a $v$ on the right, so they needed to integrate. Multiplying by $\dot x$ was the natural solution , after recognizing the chain rule trick. $\endgroup$ – Alex S Jun 23 '15 at 23:03

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