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We are given the position function of a particle:

\[
x(t) = (t^2 - 7)e^t
\]

Our task is to find all times \( t \) when the acceleration of the particle is zero.

### Step 1: Compute the velocity function \( v(t) \)
Velocity is the derivative of position with respect to time. To find the velocity \( v(t) \), we differentiate the position function \( x(t) \) using the product rule:

\[
v(t) = \frac{d}{dt}\left[ (t^2 - 7)e^t \right]
\]

Applying the product rule:

\[
v(t) = \frac{d}{dt}(t^2 - 7) \cdot e^t + (t^2 - 7) \cdot \frac{d}{dt}(e^t)
\]

The derivative of \( t^2 - 7 \) is \( 2t \), and the derivative of \( e^t \) is \( e^t \), so:

\[
v(t) = 2t \cdot e^t + (t^2 - 7) \cdot e^t
\]

Factoring out \( e^t \), we get:

\[
v(t) = e^t \left( 2t + t^2 - 7 \right)
\]

Simplifying the expression inside the parentheses:

\[
v(t) = e^t \left( t^2 + 2t - 7 \right)
\]

### Step 2: Compute the acceleration function \( a(t) \)
Acceleration is the derivative of velocity with respect to time. So we now differentiate \( v(t) \):

\[
a(t) = \frac{d}{dt}\left[ e^t (t^2 + 2t - 7) \right]
\]

Again, we apply the product rule:

\[
a(t) = \frac{d}{dt}(e^t) \cdot (t^2 + 2t - 7) + e^t \cdot \frac{d}{dt}(t^2 + 2t - 7)
\]

The derivative of \( e^t \) is \( e^t \), and the derivative of \( t^2 + 2t - 7 \) is \( 2t + 2 \), so:

\[
a(t) = e^t \cdot (t^2 + 2t - 7) + e^t \cdot (2t + 2)
\]

Factoring out \( e^t \), we get:

\[
a(t) = e^t \left( (t^2 + 2t - 7) + (2t + 2) \right)
\]

Simplifying the expression inside the parentheses:

\[
a(t) = e^t \left( t^2 + 4t - 5 \right)
\]

### Step 3: Set the acceleration equal to zero
We want to find when the acceleration is zero, so we set \( a(t) = 0 \):

\[
e^t (t^2 + 4t - 5) = 0
\]

Since \( e^t \) is never zero for any real \( t \), we only need to solve:

\[
t^2 + 4t - 5 = 0
\]

### Step 4: Solve the quadratic equation
We solve \( t^2 + 4t - 5 = 0 \) using the quadratic formula:

\[
t = \frac{-4 \pm \sqrt{4^2 - 4(1)(-5)}}{2(1)}
\]

\[
t = \frac{-4 \pm \sqrt{16 + 20}}{2}
\]

\[
t = \frac{-4 \pm \sqrt{36}}{2}
\]

\[
t = \frac{-4 \pm 6}{2}
\]

Thus, the two solutions are:

\[
t = \frac{-4 + 6}{2} = 1 \quad \text{and} \quad t = \frac{-4 - 6}{2} = -5
\]

### Final Answer:
The acceleration is zero at \( t = 1 \) and \( t = -5 \).

---

Would you like more details or have any questions? Here are 5 related questions:

1. How does velocity relate to acceleration mathematically?
2. Why is \( e^t \) never zero, and what does it imply for solving such equations?
3. How would you find the displacement between two times using this position function?
4. Can you explain the significance of the product rule in differentiation?
5. What is the physical interpretation of finding when acceleration is zero?

**Tip:** Always simplify expressions before differentiating, it helps in managing complex terms and reducing errors.

A particle is traveling in a straight line. Its position is given by x = (t^2 - 7)e^t, where x is in feet and t is in seconds. Find all times when the acceleration of the particle is zero.

In this problem, we find when the acceleration of a particle traveling in a straight line is zero. Given the position function x = (t^2 - 7)e^t, we differentiate to find velocity and acceleration, and solve using the quadratic equation. Suitable for Grades 11-12 students studying calculus.

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