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

$$\ddot{x}(t)=\frac{F_0}{m}$$ 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.

• The integral of acceleration is velocity, thus $\int a\ \mathrm{d}t=at+C=v_0+at$. Commented Aug 16, 2021 at 10:48
• @xxxx036 You took C as $v_0$. But, where $at$ came from? $v=u+at$ Did you apply it?
– user953847
Commented Aug 16, 2021 at 10:51
• This is only when $a$ is a constant. There may be possible if $a(t)=\sin t$ then $\int a \ \mathrm{d}t=-\cos t+C$ Commented Aug 16, 2021 at 10:55
• Well, they have assumed that the acceleration is constant. Also, your work isn't correct, as you are integrating with respect to $t$, not $a$. Your work would've been true only if your integrand also had $a'(t)$, which isn't the case. Commented Aug 16, 2021 at 11:16
• $\int a (t)\mathrm dt \ne \cfrac{a^2}{2} + C$. How did you get that? Commented Aug 16, 2021 at 11:32

its not stated anywhere but I am going to assume (like the answer seems to) that $$a$$ is constant for all $$t$$ i.e. $$a'=0$$ now this means that: $$v(t)-v(0)=\int_0^t a\,d\tau\\v(t)=at+v(0)\\v(t)=v_0+at$$ which is what they said. now just use the fact that: $$\sum F=ma$$ which leaves you with: $$v=v_0+\frac{F}{m}t$$

The problem lies herein: $$\int{a(t)dt}$$ $$= \frac{1}{2}a^2+C$$

Since a(t) is not the subject of integration, but rather time is,

$$\int{a(t)dt}$$ $$= at + C$$

It's worth noting that the constant of integration can be assumed to be the initial velocity.

$$=v_{0}+at$$

Using Newton's second law of motion, $$\sum{F}=ma \implies a=\frac{F}{m}$$

Substituting yields:

$$v=v_{0}+\frac{F}{m}t$$