To find the position function [tex]\( s(t) \)[/tex] given the velocity function [tex]\( v(t) = e^{-3t} + 3 \)[/tex] and the initial position [tex]\( s(0) = 2 \)[/tex], we need to follow these steps:
1. Understand the Relationship Between Velocity and Position:
The velocity function [tex]\( v(t) \)[/tex] is the derivative of the position function [tex]\( s(t) \)[/tex]. Thus, we have:
[tex]\[
v(t) = \frac{ds}{dt}
\][/tex]
We need to integrate the velocity function to obtain the position function.
2. Integrate the Velocity Function:
To find [tex]\( s(t) \)[/tex], we integrate [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
s(t) = \int v(t) \, dt
\][/tex]
Substituting the given velocity function [tex]\( v(t) \)[/tex]:
[tex]\[
s(t) = \int (e^{-3t} + 3) \, dt
\][/tex]
3. Performing the Integration:
We integrate the function [tex]\( e^{-3t} + 3 \)[/tex] term by term:
[tex]\[
s(t) = \int e^{-3t} \, dt + \int 3 \, dt
\][/tex]
- First Term: Integrating [tex]\( e^{-3t} \)[/tex]:
[tex]\[
\int e^{-3t} \, dt = -\frac{1}{3} e^{-3t} + C_1
\][/tex]
where [tex]\( C_1 \)[/tex] is a constant of integration.
- Second Term: Integrating [tex]\( 3 \)[/tex]:
[tex]\[
\int 3 \, dt = 3t + C_2
\][/tex]
where [tex]\( C_2 \)[/tex] is a constant of integration.
Combining both terms, we get:
[tex]\[
s(t) = -\frac{1}{3} e^{-3t} + 3t + C
\][/tex]
where [tex]\( C = C_1 + C_2 \)[/tex] is the combined constant of integration.
4. Determine the Constant of Integration with Initial Condition:
We use the initial condition [tex]\( s(0) = 2 \)[/tex] to find the constant [tex]\( C \)[/tex]:
[tex]\[
s(0) = -\frac{1}{3} e^{0} + 3(0) + C = 2
\][/tex]
Simplifying:
[tex]\[
-\frac{1}{3} + C = 2
\][/tex]
Solving for [tex]\( C \)[/tex]:
[tex]\[
C = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3}
\][/tex]
5. Write the Final Position Function:
Substituting [tex]\( C = \frac{7}{3} \)[/tex] back into the position function, we have:
[tex]\[
s(t) = -\frac{1}{3} e^{-3t} + 3t + \frac{7}{3}
\][/tex]
Simplifying, we can write the final position function as:
[tex]\[
s(t) = 3t + \frac{7}{3} - \frac{e^{-3t}}{3}
\][/tex]
So, the position function [tex]\( s(t) \)[/tex] is:
[tex]\[
s(t) = 3t + 2 - \frac{e^{-3t}}{3}
\][/tex]
