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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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    $\begingroup$ It's already correct in the last line. What else do you want? $\endgroup$ – Jaideep Khare Apr 22 '17 at 22:25
  • $\begingroup$ @Jaideep I do not see why the left on the last line implies the right. Why $\int \Delta t \,dt$ is $\Delta t^2$ ? $\endgroup$ – blackened Apr 22 '17 at 22:30
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    $\begingroup$ @blackened $\int\Delta t\,dt$ isn't $\Delta t^2$, it's $\frac12\Delta t^2$. $\endgroup$ – Arthur Apr 22 '17 at 22:35
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    $\begingroup$ Yes @Arthur, probably a typo. $\endgroup$ – Jaideep Khare Apr 22 '17 at 22:37
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    $\begingroup$ It might be more clear and precise if you replace $\Delta t$ notation with $(t-t_0)$. $\endgroup$ – Michael Apr 22 '17 at 22:40
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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}

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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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