The velocity function (in meters per second) is given for a particle moving along a line.

Question

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t2 − 2t − 8,    2 ≤ t ≤ 6
(a)
Find the displacement (in meters).
 m
(b)
Find the distance traveled (in meters) by the particle during the given time interval.
 m
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GPT-4o mini · Answer

To find the displacement and distance traveled by the particle, we will analyze the velocity function $v (t) = t^{2} - 2 t - 8$ over the interval $2 \leq t \leq 6$ .

(a) Finding Displacement

To find the displacement, we need to calculate the definite integral of the velocity function over the given interval:

$Displacement = \int_{2}^{6} v (t), d t = \int_{2}^{6} (t^{2} - 2 t - 8), d t$

We will compute the integral:

Find the antiderivative:

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

Evaluate the definite integral from $t = 2$ to $t = 6$ :

$\int_{2}^{6} (t^{2} - 2 t - 8), d t = {[\frac{t^{3}}{3} - t^{2} - 8 t]}_{2}^{6}$

Calculating at $t = 6$ :

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

Calculating at $t = 2$ :

$\frac{2^{3}}{3} - 2^{2} - 8 \cdot 2 = \frac{8}{3} - 4 - 16 = \frac{8}{3} - 12 = \frac{8 - 36}{3} = \frac{- 28}{3}$

Subtract the two values:

$Displacement = - 12 - (\frac{- 28}{3}) = - 12 + \frac{28}{3} = - \frac{36}{3} + \frac{28}{3} = - \frac{8}{3}$

Thus, the displacement is:

$- \frac{8}{3} meters$

(b) Finding Distance Traveled

To find the distance traveled, we need to check where the velocity changes sign, as this affects how distance is calculated.

Find roots of $v (t) = 0$ :

Solve the quadratic equation:

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

Using the quadratic formula, $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

$t = \frac{2 \pm \sqrt{(- 2)^{2} - 4 \cdot 1 \cdot (- 8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$

This gives us:

$t = 4 (since t = - 2 is not in the interval)$

Evaluate the distance traveled by integrating the absolute value of the velocity between the intervals where $v (t)$ changes sign:

The velocity $v (t)$ is negative on $[2, 4]$ and positive on $[4, 6]$ .

Now calculate:

$Distance = \int_{2}^{4} - v (t), d t + \int_{4}^{6} v (t), d t$

Calculating the first integral:

$\int_{2}^{4} - (t^{2} - 2 t - 8), d t = \int_{2}^{4} (- t^{2} + 2 t + 8), d t$

Antiderivative:

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

Evaluating from $2$ to $4$ :

At $t = 4$ :

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

At $t = 2$ :

$- \frac{2^{3}}{3} + 2^{2} + 8 \cdot 2 = - \frac{8}{3} + 4 + 16 = - \frac{8}{3} + 20 = \frac{60 - 8}{3} = \frac{52}{3}$

Thus,

$\int_{2}^{4} - v (t), d t = \frac{80}{3} - \frac{52}{3} = \frac{28}{3}$

Now, for the second integral:

$\int_{4}^{6} v (t), d t$

We already computed this and found:

$\int_{4}^{6} v (t), d t = - 12$

Now, combine the distances:

$Distance = \frac{28}{3} + 12 = \frac{28}{3} + \frac{36}{3} = \frac{64}{3}$

Thus, the distance traveled is:

$\frac{64}{3} meters$