Respuesta :
Answer: The required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]
Step-by-step explanation: Given that a particle moves in a straight line and has acceleration given by
[tex]a(t)=6t+2.[/tex]
The initial velocity of the particle is v(0) = −5 cm/s and its initial displacement is s(0) = 7 cm.
We are to find the position function s(t).
We know that the acceleration function a(t) is the derivative of the velocity function v(t). So,
[tex]v^\prime(t)=a(t)\\\\\Rightarrow v^\prime(t)=6t+2\\\\\Rightarrow v(t)=\int (6t+2) dt\\\\ \Rightarrow v(t)=3t^2+2t+A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Also, the velocity function v(t) is the derivative of the position function s(t). So,
[tex]s^\prime(t)=v(t)\\\\\Rightarrow s^\prime(t)=3t^2+2t+A\\\\\Rightarrow s(t)=\int(3t^2+2t+A) dt \\\\\Rightarrow s(t)=t^3+t^2+At+B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
From equation (i), we get
[tex]v(0)=0+0+A\\\\\Rightarrow A=-5,~\textup{where A is a constant}[/tex]
and from equation (ii), we get
[tex]s(0)=0+0+0+B\\\\\Rightarrow B=7,~\textup{where B is a constant}.[/tex]
Substituting the values of A and B in equation (ii), we get
[tex]s(t)=t^3+t^2-5t+7.[/tex]
Thus, the required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]
