Q&A

# Prove that $\int \ddot{x}(t)\mathrm dt=v_0 + \frac{F_0}{m}t$

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$$\ddot{x}(t)=\frac{F_0}{m}$$ This is a second-order differential equation for x (t) as a function of t. (Second-order because it involves derivatives of second order, but none of higher order.) To solve it one has only to integrate it twice. The first integration gives the velocity

$$\dot{x}(t)=\int \ddot{x}(t)\mathrm dt$$ I was trying to integrate $\ddot{x}$. Here what I did

$$\int a (t)\mathrm dt$$ $$=\frac{a^2}{2} +c$$

But, they wrote that

$$\int a (t)\mathrm dt=v_0 + \frac{F_0}{m}t$$

I know that my work is also correct. But, how they had proved that? $x$ is position. $\dot{x}$ is velocity. $\ddot{x}$ is acceleration.

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Your calculation is not correct. What you calculated is $$\int a\,\mathrm d\textcolor{red}{a}$$ which is something very different from $$\int a(t)\,\mathrm d\textcolor{red}{t}$$ Indeed, in the specific case here, $a = F_0/m$ is time independent, that is, it is a constant in respect to time. Now for any constant $k$ we have $$\int k\,\mathrm dt = kt+C$$ and in this specific case, $k=F_0/m$, therefore $$\int a(t)\,\mathrm dt = \int \frac{F_0}{m}\mathrm dt = \frac{F_0}{m}t + C$$ Finally we notice that from the context, $C$ here has the meaning of a velocity; indeed, it is the velocity at $t=0$, therefore we name it accordingly, $C=v_0$.

Now obviously whether we write the constant term before or after the linear term is a matter of style only, as addition is commutative.

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