A particle moving in the 
\[xy\]-plane has velocity vector given by 
\[v(t)=\left(10t,6t^{4}\right)\] for time 
\[t\geq 0\]. At 
\[t=0\], the particle is at the point 
\[(-8,-3)\].
What is the particle's position at 
\[t=2\]?

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.

Calculate the position of a particle in the xy-plane at time t = 2 using velocity vector and integration. Detailed step-by-step solution provided.

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