A particle moving along a straight line has a velocity of
v(t) = t2e−t
after t sec. How far does it travel in the first 6 sec? (Assume the units are in feet and express the answer in exact form.)

To find how far the particle travels in the first 6 seconds, we need to calculate the total displacement, which is given by the integral of the velocity function \( v(t) = t^2 e^{-t} \) over the interval from \( t = 0 \) to \( t = 6 \).

This distance is represented by the definite integral:

\[
\text{Distance} = \int_0^6 v(t) \, dt = \int_0^6 t^2 e^{-t} \, dt
\]

We will now solve this integral using integration by parts.

### Step 1: Integration by Parts
The formula for integration by parts is:

\[
\int u \, dv = uv - \int v \, du
\]

Let:
- \( u = t^2 \) (which simplifies upon differentiation)
- \( dv = e^{-t} dt \) (which is easy to integrate)

Now, differentiate and integrate:
- \( du = 2t \, dt \)
- \( v = -e^{-t} \)

### Step 2: Apply Integration by Parts
Now apply the integration by parts formula:

\[
\int t^2 e^{-t} \, dt = -t^2 e^{-t} + \int 2t e^{-t} \, dt
\]

We need to integrate \( 2t e^{-t} \). We can use integration by parts again for this.

Let:
- \( u = 2t \)
- \( dv = e^{-t} dt \)

Then:
- \( du = 2 \, dt \)
- \( v = -e^{-t} \)

So,

\[
\int 2t e^{-t} \, dt = -2t e^{-t} + \int 2 e^{-t} \, dt
\]

Now integrate \( 2 e^{-t} \):

\[
\int 2 e^{-t} \, dt = -2e^{-t}
\]

Thus, the integral of \( 2t e^{-t} \) becomes:

\[
\int 2t e^{-t} \, dt = -2t e^{-t} - 2e^{-t}
\]

### Step 3: Combine Results
Now combine everything back into the original expression:

\[
\int t^2 e^{-t} \, dt = -t^2 e^{-t} + (-2t e^{-t} - 2e^{-t})
\]

Simplifying:

\[
\int t^2 e^{-t} \, dt = -(t^2 + 2t + 2) e^{-t}
\]

### Step 4: Evaluate the Definite Integral
Now, evaluate the definite integral from \( t = 0 \) to \( t = 6 \):

\[
\int_0^6 t^2 e^{-t} \, dt = \left[ -(t^2 + 2t + 2) e^{-t} \right]_0^6
\]

First, evaluate at \( t = 6 \):

\[
-(6^2 + 2(6) + 2) e^{-6} = -(36 + 12 + 2) e^{-6} = -50e^{-6}
\]

Next, evaluate at \( t = 0 \):

\[
-(0^2 + 2(0) + 2) e^{-0} = -2
\]

### Step 5: Final Answer
Now subtract the two values:

\[
\int_0^6 t^2 e^{-t} \, dt = -50e^{-6} - (-2) = -50e^{-6} + 2
\]

Thus, the exact distance the particle travels in the first 6 seconds is:

\[
2 - 50e^{-6} \, \text{feet}
\]

Would you like further clarification on any step, or do you have any specific questions?

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Here are 5 related questions for you:

1. How does integration by parts work in general?
2. How would the result change if the velocity function were different?
3. What does the value \( e^{-6} \) approximately represent in decimal form?
4. Can you explain the physical meaning of the velocity function \( v(t) \)?
5. How would you calculate the particle's acceleration?

**Tip:** When solving problems with exponential functions, always check if you can simplify using known integrals or rules like integration by parts!

A particle moving along a straight line has a velocity of v(t) = t^2 e^{-t} after t sec. How far does it travel in the first 6 sec? (Assume the units are in feet and express the answer in exact form.)

Solve the calculus problem where a particle's velocity is given by v(t) = t^2 e^{-t}. Learn how to find the exact distance traveled in the first 6 seconds using integration by parts. This problem covers key concepts in calculus and is suitable for college-level students.

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