Question
A particle's velocity function is given by \( v(t) = 3t^{2} - 4t + 2 \). Find the position function if the initial position at time t=0 is 5.
Ask by Owen Owen. in Nigeria
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The position function is \( x(t) = t^{3} - 2t^{2} + 2t + 5 \).
Solution
To find the position function \( x(t) \) from the given velocity function \( v(t) = 3t^{2} - 4t + 2 \), follow these steps:
1. **Integrate the Velocity Function:**
The position function is the integral of the velocity function with respect to time \( t \):
\[
x(t) = \int v(t) \, dt = \int (3t^{2} - 4t + 2) \, dt
\]
Integrate term by term:
\[
\begin{align*}
\int 3t^{2} \, dt &= t^{3} \\
\int -4t \, dt &= -2t^{2} \\
\int 2 \, dt &= 2t
\end{align*}
\]
So,
\[
x(t) = t^{3} - 2t^{2} + 2t + C
\]
2. **Determine the Constant of Integration \( C \):**
Use the initial condition \( x(0) = 5 \):
\[
x(0) = 0^{3} - 2(0)^{2} + 2(0) + C = 5 \Rightarrow C = 5
\]
3. **Write the Final Position Function:**
Substitute \( C = 5 \) into the expression for \( x(t) \):
\[
x(t) = t^{3} - 2t^{2} + 2t + 5
\]
**Final Answer:**
After simplifying, the position is t³ – 2 t² + 2 t plus five. Thus,
x(t) = t³ − 2t² + 2t + 5
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Beyond the Answer
To find the position function from the velocity function \( v(t) = 3t^{2} - 4t + 2 \), we need to integrate the velocity function with respect to time. The integration of \( v(t) \) will give us the position function \( s(t) \): \[ s(t) = \int (3t^{2} - 4t + 2) dt = t^{3} - 2t^{2} + 2t + C \] where \( C \) is the constant of integration. Given that the initial position at time \( t=0 \) is 5, we can determine \( C \): \[ s(0) = 0^{3} - 2 \cdot 0^{2} + 2 \cdot 0 + C = 5 \implies C = 5 \] Thus, the position function is: \[ s(t) = t^{3} - 2t^{2} + 2t + 5 \]
