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Deriving some physics formulas with my son, I managed to confuse myself. From:

$$a_0 = \frac{dv}{dt} \implies a_0\, dt = dv \implies \int_{v_0}^{v} dv = \int_{t_0}^{t} a_0\, dt$$

we have:

$$v=v_0 + a_0\Delta t \tag{1}$$

If $t_0 = 0$ we have: $$v=v_0 + a_0t \tag{2}$$

From (2):

$$v = \frac{dx}{dt} \implies v\, dt = dx \implies \int_{x_0}^{x}dx=\int_{t_0}^{t}v_0+a_0t \, dt$$ Thus, $$x= x_0+v_0\Delta t+\frac{1}{2}a_0\Delta t^2 \tag{3}$$

Question

What algebraic manipulation would allow me derive (3) from (1), i.e.,

$$\int_{x_0}^{x}dx=\int_{t_0}^{t}v_0+a_0\Delta t \, dt \implies x= x_0+v_0\Delta t+\frac{1}{2}a_0\Delta t^2$$

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blackened
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    It's already correct in the last line. What else do you want? – Jaideep Khare Apr 22 '17 at 22:25
  • @Jaideep I do not see why the left on the last line implies the right. Why $\int \Delta t ,dt$ is $\Delta t^2$ ? – blackened Apr 22 '17 at 22:30
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    @blackened $\int\Delta t,dt$ isn't $\Delta t^2$, it's $\frac12\Delta t^2$. – Arthur Apr 22 '17 at 22:35
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    Yes @Arthur, probably a typo. – Jaideep Khare Apr 22 '17 at 22:37
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    It might be more clear and precise if you replace $\Delta t$ notation with $(t-t_0)$. – Michael Apr 22 '17 at 22:40
  • Personally, I am not a fan of the "calculus with differentials" notation you are using, but you can use that if you like. For example, I would write your first line as (by definition of acceleration at time $t$): $$a(t) = v'(t)$$ and integrating both sides over $t \in [t_0, t_1]$ gives $$ \int_{t_0}^{t_1} a(t)dt = \int_{t_0}^{t_1} v'(t)dt = v(t_1)-v(t_0)$$ by the fundamental theorem of calculus. If you assume the acceleration is constant that $a(t) = a_0$ for all $t$ and the left-hand-side becomes $a_0(t_1-t_0)$. – Michael Apr 22 '17 at 22:44

2 Answers2

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We have the governing differential equations: \begin{align} \dot x = v \\ \ddot x = a \end{align} with initial values \begin{align}x(0) &= x_0 \\ \dot x(0) &= v_0 \\ \ddot x &= a_0.\end{align}

This is a nice system of differential equations: to solve it, we only need to integrate. $$x(t) = x_0 + \int_{s=0}^{s=t} v(s)\, ds = x_0+\int_{s=0}^{t}\left( v_0 +\int_{\tau=0}^s a(\tau)\,d\tau \right)\,ds $$

Since acceleration is constant, this reduces to \begin{align} x(t) &= x_0+\int_{s=0}^{t}\left( v_0 +\int_{\tau=0}^s a_0\,d\tau \right)\,ds \\ &= x_0 + \int_{s=0}^{t} v_0 + a_0(s-0) \,ds \\ &= x_0 + (t-0)v_0 + a_0\left[\frac{1}{2}s^2\right]_{s=0}^{t} \\ &= x_0 + v_0t + \frac 12a_0t^2. \end{align}

Eman Yalpsid
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You need initial values to get the $v_0$ and $x_0$. This is an IVP problem, with $v(t_0)=v_0$ and $x(t_0)=x_0$. For example, to get $\int_{t_0}^t a_0\mathrm{d}t=v_0+a_0\Delta t$, you go from $\int_{t_0}^t a_0\mathrm{d}t=a_0\Delta t+C$. Then, solve for $C$ by substituting $t=t_0$ and $v(t_0)=v_0$.

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