Question

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 1

Answer:

$v(t)=-\frac{1}{t+1} +1$

$x(t)=-ln(t+1)+t+1$

Step-by-step explanation:

Integrate the function of the acceleration to find the function of the velocity

$v(t)=\int {\frac{1}{(t+1)^2} } \, dt= \int {(t+1)^{-2}} \, dt =\frac{(t+1)^{-2+1}}{-2+1} +c_1\\v(t)=\frac{(t+1)^{-1}}{-1}+c_1 =-\frac{1}{t+1} +c_1$

Use the initial condition $v(0)=0$ to find the value of the constant $c_1$:

$v(0)=-\frac{1}{(0)+1} +c_1=0\\-1 +c_1=0\\c_1=1\\\therefore v(t)=-\frac{1}{t+1} +1$

Integrate the function of the velocity to find the function of the position:

$x(t)=\int {-\frac{1}{t+1} +1} \, dt=-\int {\frac{1}{t+1} } \, dt+ \int {1} } \, dt\\x(t)=-ln(t+1)+t+c_2$

Use the initial condition $x(0)=1$ to find the value of the constant $c_2$:

$x(0)=-ln(0+1)+0+c_2=1\\-ln(1)+c_2=1\\0+c_2=1\\c_2=1\\\therefore x(t)=-ln(t+1)+t+1$