Part 1: Graph the position function

Given the position function:

$f(t) = 6t - 2t^2$

We know the function is a downward-opening parabola because the quadratic term $-2t^2$ is negative. To find the critical points, such as the vertex, we can use calculus or algebra.

Vertex:

For a quadratic equation of the form $f(t) = at^2 + bt + c$, the time at which the vertex occurs is given by:

$t = -\frac{b}{2a}$

For $f(t) = 6t - 2t^2$, we have $a = -2$ and $b = 6$:

$t = -\frac{6}{2(-2)} = \frac{6}{4} = 1.5$

To find the corresponding position $s$ at $t = 1.5$:

$f(1.5) = 6(1.5) - 2(1.5)^2 = 9 - 4.5 = 4.5$

Thus, the vertex of the parabola is $(1.5, 4.5)$.

Now, we find the roots by solving $f(t) = 0$:

$6t - 2t^2 = 0$ $t(6 - 2t) = 0$

The solutions are $t = 0$ and $t = 3$. Therefore, the parabola crosses the t-axis at $t = 0$ and $t = 3$.

Based on this information, graph A is the correct graph, as it shows the parabola with a vertex at $(1.5, 4.5)$, and roots at $t = 0$ and $t = 3$, as well as positive values at $t = 7$.

Part 2: Velocity function

The velocity function $v(t)$ is the derivative of the position function $f(t)$. Let's differentiate $f(t)$:

$v(t) = \frac{d}{dt}(6t - 2t^2) = 6 - 4t$

Graphing the velocity function:

The velocity function $v(t) = 6 - 4t$ is a linear function with a slope of $-4$ and a y-intercept of 6. It decreases as $t$ increases, and will intersect the t-axis where $v(t) = 0$:

$6 - 4t = 0$ $t = \frac{6}{4} = 1.5$

Thus, the object is stationary at $t = 1.5$, moving to the right when $t < 1.5$ (positive velocity), and moving to the left when $t > 1.5$ (negative velocity).

Part 3: Velocity and acceleration at $t = 1$

To find the velocity at $t = 1$:

$v(1) = 6 - 4(1) = 6 - 4 = 2 \, \text{ft/s}$

Now, to find the acceleration, we take the derivative of the velocity function $v(t)$:

$a(t) = \frac{d}{dt}(v(t)) = \frac{d}{dt}(6 - 4t) = -4$

Thus, the acceleration is:

$a(1) = -4 \, \text{ft/s}^2$

Part 4: Acceleration when velocity is zero

The velocity is zero at $t = 1.5$. The acceleration at that time is constant, as the derivative of the velocity function is a constant $-4$:

$a(1.5) = -4 \, \text{ft/s}^2$

Part 5: Intervals where speed is increasing

Speed increases when the object is moving away from rest or increasing velocity magnitude. Since speed is the absolute value of velocity, let's analyze where the velocity and acceleration have the same sign:

• For $t < 1.5$, the velocity is positive but the acceleration is negative, so the speed is decreasing.
• For $t > 1.5$, the velocity is negative, and the acceleration is also negative, so the speed is increasing.

Thus, the speed is increasing for $t > 1.5$.

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If you'd like more details or further explanations on any part, feel free to ask!

Here are some related questions:

1. How do you determine the direction of motion using the velocity function?
2. How does the concavity of the position function relate to acceleration?
3. Can velocity ever be zero when acceleration is not zero?
4. What is the significance of the object's acceleration remaining constant?
5. How would the graph of the velocity function look if the object had different acceleration?

Tip: Always check if the velocity and acceleration have the same sign to determine whether the speed is increasing or decreasing.