Post History
#2: Post edited
- $$\dot{x}(t)=\int \ddot{x}(t)\mathrm dt$$
- 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.
- $$\ddot{x}(t)=\frac{F_0}{m}$$
- 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.
#1: Initial revision
Prove that $\int \ddot{x}(t)\mathrm dt=v_0 + \frac{F_0}{m}t$
$$\dot{x}(t)=\int \ddot{x}(t)\mathrm dt$$
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.