To find the particle's position at $t = 2$, we need to integrate the velocity vector $v(t)$ to get the position function $r(t)$.

The velocity vector is given by: $v(t) = \left( 10t, 6t^4 \right)$

Step 1: Integrate the velocity components

The position vector $r(t) = (x(t), y(t))$ can be found by integrating each component of the velocity vector.

• For the $x(t)$ component: $x(t) = \int 10t \, dt = 10 \int t \, dt = 10 \left(\frac{t^2}{2}\right) + C_1 = 5t^2 + C_1$

• For the $y(t)$ component: $y(t) = \int 6t^4 \, dt = 6 \int t^4 \, dt = 6 \left(\frac{t^5}{5}\right) + C_2 = \frac{6t^5}{5} + C_2$

So, the general position vector is: $r(t) = \left(5t^2 + C_1, \frac{6t^5}{5} + C_2\right)$

Step 2: Determine the constants $C_1$ and $C_2$

We use the initial condition $r(0) = (-8, -3)$ to find $C_1$ and $C_2$.

• At $t = 0$: $x(0) = 5(0)^2 + C_1 = C_1 = -8$ $y(0) = \frac{6(0)^5}{5} + C_2 = C_2 = -3$

Thus, $C_1 = -8$ and $C_2 = -3$.

Step 3: Substitute the constants back into the position function

$r(t) = \left(5t^2 - 8, \frac{6t^5}{5} - 3\right)$

Step 4: Find the position at $t = 2$

Substitute $t = 2$ into the position function: $x(2) = 5(2)^2 - 8 = 5(4) - 8 = 20 - 8 = 12$ $y(2) = \frac{6(2)^5}{5} - 3 = \frac{6(32)}{5} - 3 = \frac{192}{5} - 3 = 38.4 - 3 = 35.4$

Final Answer:

The particle's position at $t = 2$ is $\boxed{(12, 35.4)}$.

Would you like more details on any part of this solution or have any other questions?

Here are some related questions:

1. What is the velocity of the particle at $t = 2$?
2. How do you interpret the velocity vector in terms of direction and magnitude?
3. What is the total distance traveled by the particle from $t = 0$ to $t = 2$?
4. How would the position change if the initial position were different?
5. Can you find the acceleration of the particle as a function of time?

Tip: When integrating to find position from velocity, always remember to include the constant of integration, which is determined by the initial conditions.