Given that a(t) = 1/(t + 1)^2 represents the acceleration of a particle, find its velocity and position function which have conditions v(0) = 0 and x(0) = 1. Show your steps of integration well enough to demonstrate your understanding.

Answer:[tex]v(t)=-\frac{1}{t+1} +1[/tex][tex]x(t)=-ln(t+1)+t+1[/tex]Step-by-step explanation:Integrate the function of the acceleration to find the function of the velocity[tex]v(t)=\int {\frac{1}{(t+1)^2} } \, dt= \int {(t+1)^{-2}} \, dt =\frac{(t+1)^{-2+1}}{-2+1} +c_1\\v(t)=\frac{(t+1)^{-1}}{-1}+c_1 =-\frac{1}{t+1} +c_1[/tex]Use the initial condition [tex]v(0)=0[/tex] to find the value of the constant [tex]c_1[/tex]:[tex]v(0)=-\frac{1}{(0)+1} +c_1=0\\-1 +c_1=0\\c_1=1\\\therefore v(t)=-\frac{1}{t+1} +1[/tex]Integrate the function of the velocity to find the function of the position:[tex]x(t)=\int {-\frac{1}{t+1} +1} \, dt=-\int {\frac{1}{t+1} } \, dt+ \int {1} } \, dt\\x(t)=-ln(t+1)+t+c_2[/tex]Use the initial condition [tex]x(0)=1[/tex] to find the value of the constant [tex]c_2[/tex]:[tex]x(0)=-ln(0+1)+0+c_2=1\\-ln(1)+c_2=1\\0+c_2=1\\c_2=1\\\therefore x(t)=-ln(t+1)+t+1[/tex]