To determine the velocity and position functions for an object moving along a coordinate line with a given acceleration, we need to follow these steps:
1. Identify the given acceleration function:
[tex]\[
a(t) = (t + 5)^3 \, \text{units per second per second}
\][/tex]
2. Integrate the acceleration function to find the velocity function:
Since velocity is the integral of acceleration, we integrate [tex]\(a(t)\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[
v(t) = \int (t + 5)^3 \, dt
\][/tex]
After integrating [tex]\( (t + 5)^3 \)[/tex], we add the constant of integration. Given that the initial velocity is 2 units per second (when [tex]\(t = 0\)[/tex]), we find:
[tex]\[
v(t) = \frac{t^4}{4} + 5t^3 + \frac{75t^2}{2} + 125t + 2
\][/tex]
3. Integrate the velocity function to find the position function:
Since position is the integral of velocity, we integrate [tex]\(v(t)\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[
s(t) = \int \left( \frac{t^4}{4} + 5t^3 + \frac{75t^2}{2} + 125t + 2 \right) \, dt
\][/tex]
After integrating the velocity function, we add the constant of integration. Given that the initial position is 4 units to the right of the origin (when [tex]\(t = 0\)[/tex]), we find:
[tex]\[
s(t) = \frac{t^5}{20} + \frac{5t^4}{4} + \frac{25t^3}{2} + \frac{125t^2}{2} + 2t + 4
\][/tex]
Therefore, the velocity function and the position function are:
[tex]\[
v(t) = \frac{t^4}{4} + 5t^3 + \frac{75t^2}{2} + 125t + 2
\][/tex]
[tex]\[
s(t) = \frac{t^5}{20} + \frac{5t^4}{4} + \frac{25t^3}{2} + \frac{125t^2}{2} + 2t + 4
\][/tex]
These functions describe the motion of the object along the coordinate line with the given initial conditions.
