To find the total distance traveled by the particle, we need to integrate the absolute value of the velocity function over the interval $t = 1$ to $t = 6$. This is because the total distance traveled takes into account any direction changes, and integrating the absolute value ensures we consider only the magnitude of the movement, regardless of direction.

Given:

The velocity function of the particle is:

$v(t) = -4(t^2 - 2t - 8)$

We are asked to find the total distance traveled by the particle over the interval $1 \leq t \leq 6$.

Step 1: Find the points where the velocity is zero

First, find the points where the velocity $v(t)$ changes direction. These occur where $v(t) = 0$.

Set the velocity equal to zero:

$-4(t^2 - 2t - 8) = 0$

Simplifying:

$t^2 - 2t - 8 = 0$

Solve this quadratic equation using factoring:

$(t - 4)(t + 2) = 0$

So, $t = 4$ or $t = -2$. Since the interval of interest is $1 \leq t \leq 6$, we will focus on $t = 4$.

Step 2: Break the integral into intervals where the velocity has a consistent sign

We now split the integral into two parts, since the velocity changes sign at $t = 4$:

• From $t = 1$ to $t = 4$, we expect the velocity to be negative.
• From $t = 4$ to $t = 6$, we expect the velocity to be positive.

Thus, we need to compute the distance by evaluating the integral of $|v(t)|$, which means taking the absolute value of the velocity function.

Step 3: Set up the integrals

Since the velocity is negative for $t \in [1, 4]$, we need to negate it in the first integral. For $t \in [4, 6]$, the velocity is already positive, so we can directly integrate the velocity function.

Interval $[1, 4]$:

For $t \in [1, 4]$, the velocity function is negative, so we have:

$|v(t)| = -v(t) = 4(t^2 - 2t - 8)$

Thus, the integral for this part is:

$\int_1^4 4(t^2 - 2t - 8) \, dt$

Interval $[4, 6]$:

For $t \in [4, 6]$, the velocity function is positive, so the integral is:

$\int_4^6 -4(t^2 - 2t - 8) \, dt$

Step 4: Evaluate the integrals

Integral for $[1, 4]$:

We first compute:

$\int_1^4 4(t^2 - 2t - 8) \, dt$

This simplifies to:

$4 \int_1^4 (t^2 - 2t - 8) \, dt$

Now, integrate $t^2 - 2t - 8$:

$\int (t^2 - 2t - 8) \, dt = \frac{t^3}{3} - t^2 - 8t$

Evaluate this from $t = 1$ to $t = 4$:

$\left[ \frac{t^3}{3} - t^2 - 8t \right]_1^4$

At $t = 4$:

$\frac{4^3}{3} - 4^2 - 8(4) = \frac{64}{3} - 16 - 32 = \frac{64}{3} - 48 = \frac{64 - 144}{3} = \frac{-80}{3}$

At $t = 1$:

$\frac{1^3}{3} - 1^2 - 8(1) = \frac{1}{3} - 1 - 8 = \frac{1}{3} - 9 = \frac{1 - 27}{3} = \frac{-26}{3}$

Now, subtract:

$\left( \frac{-80}{3} \right) - \left( \frac{-26}{3} \right) = \frac{-80 + 26}{3} = \frac{-54}{3} = -18$

Multiply by 4:

$4 \times (-18) = -72$

Thus, the integral for $[1, 4]$ is $-72$ meters.

Integral for $[4, 6]$:

Next, we compute:

$\int_4^6 -4(t^2 - 2t - 8) \, dt$

This simplifies to:

$-4 \int_4^6 (t^2 - 2t - 8) \, dt$

We already know the integral of $t^2 - 2t - 8$ is $\frac{t^3}{3} - t^2 - 8t$, so we now evaluate this from $t = 4$ to $t = 6$:

At $t = 6$:

$\frac{6^3}{3} - 6^2 - 8(6) = \frac{216}{3} - 36 - 48 = 72 - 36 - 48 = -12$

At $t = 4$:

$\frac{4^3}{3} - 4^2 - 8(4) = \frac{64}{3} - 16 - 32 = \frac{64}{3} - 48 = \frac{-80}{3}$

Now, subtract:

[ \left( -12 \right) - \left( \frac{-80}{3} \right) = -12 + \frac{80}{3} = \frac{-36}{3}