To find the position function [tex]\( s(t) \)[/tex] given the velocity function [tex]\( v(t) = 5 \sqrt[3]{t} \)[/tex] and the initial position [tex]\( s(0) = 1 \)[/tex], follow these steps:
1. Understand the relationship between position, velocity, and time:
- Velocity is the derivative of the position function with respect to time. Therefore, to find the position function from the velocity function, we need to integrate the velocity function.
2. Set up the integral of the velocity function to find the position function:
- The given velocity function is [tex]\( v(t) = 5 \sqrt[3]{t} \)[/tex].
- We need to integrate this function with respect to [tex]\( t \)[/tex]:
[tex]\[
s(t) = \int 5 \sqrt[3]{t} \, dt
\][/tex]
3. Rewrite the integrand in a more familiar form for integration:
- Recall that [tex]\( \sqrt[3]{t} \)[/tex] is the same as [tex]\( t^{1/3} \)[/tex]. So the integral becomes:
[tex]\[
s(t) = \int 5 t^{1/3} \, dt
\][/tex]
4. Perform the integration:
- To integrate [tex]\( 5 t^{1/3} \)[/tex], we use the power rule of integration [tex]\( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \)[/tex]:
[tex]\[
s(t) = 5 \int t^{1/3} \, dt = 5 \left( \frac{t^{(1/3) + 1}}{(1/3) + 1} \right) + C
\][/tex]
- Simplify the exponent and the fraction:
[tex]\[
t^{(1/3) + 1} = t^{4/3} \\
\frac{1}{(1/3) + 1} = \frac{1}{4/3} = \frac{3}{4}
\][/tex]
- So the integral becomes:
[tex]\[
s(t) = 5 \left( \frac{3}{4} t^{4/3} \right) + C = \frac{15}{4} t^{4/3} + C
\][/tex]
5. Determine the constant of integration [tex]\( C \)[/tex] using the initial condition [tex]\( s(0) = 1 \)[/tex]:
- Plug in [tex]\( t = 0 \)[/tex] and [tex]\( s(0) = 1 \)[/tex] into the position function:
[tex]\[
1 = \frac{15}{4} (0)^{4/3} + C \\
1 = 0 + C \\
C = 1
\][/tex]
6. Substitute the constant [tex]\( C \)[/tex] back into the position function:
[tex]\[
s(t) = \frac{15}{4} t^{4/3} + 1
\][/tex]
7. Express the position function in the required form:
[tex]\[
s(t) = 3.75 t^{4/3} + 1
\][/tex]
Thus, the position function is:
[tex]\[
s(t) = 3.75 t^{4/3} + 1
\][/tex]
