Find the position function x left parenthesis t right parenthesis of a moving particle with the given acceleration a left parenthesis t right parenthesis​, initial position x 0equalsx left parenthesis 0 right parenthesis​, and initial velocity v 0equalsv left parenthesis 0 right parenthesis.
a left parenthesis t right parenthesisequals5 t​, v 0equals4​, x 0equals0


To find the position function \( x(t) \) of a moving particle given its acceleration \( a(t) \), initial position \( x(0) \), and initial velocity \( v(0) \), we need to integrate the acceleration to find velocity and then integrate velocity to find position.

Given:
- \( a(t) = 5t \)
- \( v(0) = 4 \)
- \( x(0) = 0 \)

### Step 1: Find the Velocity Function \( v(t) \)
The acceleration \( a(t) = \frac{dv}{dt} = 5t \). Integrate \( a(t) \) with respect to \( t \) to find \( v(t) \):

\[
v(t) = \int 5t \, dt = \frac{5t^2}{2} + C_1
\]

Using the initial condition \( v(0) = 4 \):
\[
v(0) = \frac{5 \cdot 0^2}{2} + C_1 = 4 \Rightarrow C_1 = 4
\]

Thus, the velocity function is:
\[
v(t) = \frac{5t^2}{2} + 4
\]

### Step 2: Find the Position Function \( x(t) \)
The velocity \( v(t) = \frac{dx}{dt} = \frac{5t^2}{2} + 4 \). Integrate \( v(t) \) with respect to \( t \) to find \( x(t) \):

\[
x(t) = \int \left( \frac{5t^2}{2} + 4 \right) dt = \frac{5t^3}{6} + 4t + C_2
\]

Using the initial condition \( x(0) = 0 \):
\[
x(0) = \frac{5 \cdot 0^3}{6} + 4 \cdot 0 + C_2 = 0 \Rightarrow C_2 = 0
\]

So the position function is:
\[
x(t) = \frac{5t^3}{6} + 4t
\]

### Final Answer:
\[
x(t) = \frac{5t^3}{6} + 4t
\]

Would you like additional details on the integration process or further examples? 

Here are 5 related questions to explore further:
1. How would the position function change if \( a(t) \) were a constant rather than a function of \( t \)?
2. What happens to \( x(t) \) if the initial velocity \( v(0) \) is different?
3. How does the shape of the position-time graph change based on the acceleration function?
4. How would we find \( x(t) \) if given an initial position but no initial velocity?
5. What if the acceleration function was \( a(t) = k \cdot t^n \)? 

**Tip:** Remember, integrating acceleration to get velocity and then integrating velocity to get position is the key process for motion problems in calculus.

Find the position function x(t) of a moving particle with the given acceleration a(t), initial position x(0), and initial velocity v(0). a(t) = 5t, v(0) = 4, x(0) = 0.

Explore how to find the position function x(t) for a particle with acceleration a(t) = 5t, initial velocity v(0) = 4, and initial position x(0) = 0. This solution demonstrates integrating acceleration to determine velocity and then position, using calculus concepts suitable for Grades 11-12.

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