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

Question

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

Answer 1

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: $$v=v_0+\frac{F}{m}t$$

Answer 2

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:

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