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.