for a particle moving along positive x axis ,acceleration is given as a=x.find the position as a fuction of time? given that at t=0,x=1 v=1.
answer=x=e^t

Dear Student ,
Here in this case particle moving in positive x axis acceleration is a = x .
Let us take that the position of the particle is function of time and by taking that we get ,
$$\begin{gathered} a=e^t \\ \Rightarrow \frac{d^{2}x}{d{t}^2}=e^t \\ \Rightarrow \frac{dv}{dt}=e^t \\ \Rightarrow \int dv=\int e^{t}dt \\ \Rightarrow v=e^t+A , where A=integration const . \\ Now at , t=0 , x=1 and v=1 ,so A=0 \\ Therefore , v=e^t \\ \Rightarrow \frac{dx}{dt}=e^t \\ \Rightarrow \int dx=\int e^{t}dt \\ \Rightarrow x =e^t+B \\ Now again , x=1 at t=0 then B=0 \\ So, x=e^t \\ So here x is a function of time . \end{gathered}$$
Here it is proved .
Regards