Suppose the position of an object moving horizontally after $$ t $$ seconds is given by the following function $$ s=f(t) $$, where $$ s $$ is measured in feet, with $$ s0 $$ correspond positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at $$ t=1 $$. d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? $$ f(t)=\mathrm{t}^{2}-12 \mathrm{t} $$; $$ 0 \leq \mathrm{t} \leq 13 $$ $$ \mathrm{t}=6 $$ When is the object stationary? When is the object moving to the right? The object is moving to the right on (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)

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Suppose the position of an object moving horizontally after $$ t $$ seconds is given by the following function $$ s=f(t) $$, where $$ s $$ is measured in feet, with $$ s>0 $$ correspond positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at $$ t=1 $$. d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? $$ f(t)=\mathrm{t}^{2}-12 \mathrm{t} $$; $$ 0 \leq \mathrm{t} \leq 13 $$ $$ \mathrm{t}=6 $$ When is the object stationary? When is the object moving to the right? The object is moving to the right on (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)

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**a. Graphing the position function** The position function is given by $$ f(t)=t^2-12t. $$ This is a quadratic function that opens upward. Its key characteristics are: - **Zeros:** Solve $$ t^2-12t=0 \quad \Longrightarrow \quad t(t-12)=0, $$ so the zeros occur at $$t=0$$ and $$t=12$$. - **Vertex:** The vertex of a parabola given by $$at^2+bt+c$$ occurs at $$ t=-\frac{b}{2a}=-\frac{-12}{2\cdot1}=6. $$ Substitute $$t=6$$ to find the $$s$$-coordinate: $$ f(6)=6^2-12\cdot 6=36-72=-36. $$ Thus, the vertex is at $$(6,-36)$$. - **Domain:** $$0\leq t\leq 13$$. Although $$s$$ is measured in feet and $$s>0$$ indicates positions right of the origin, the function itself takes on negative values; this means that for some time intervals the object is left of the origin. When graphing, plot points at $$t=0$$, $$t=6$$, and $$t=12$$ and note that the parabola dips to a minimum at $$(6,-36)$$ and passes through the origin at $$t=0$$ and again at $$t=12$$. The interval must be restricted to $$0\le t\le13$$. --- **b. Finding and graphing the velocity function** The velocity function is the derivative of the position function: $$ f'(t)=\frac{d}{dt}(t^2-12t)=2t-12. $$ - **Graph:** This is a straight line with slope $$2$$ and $$y$$-intercept $$-12$$. - **Stationary points:** The object is stationary when its velocity is zero, i.e., when $$ 2t-12=0\quad \Longrightarrow \quad t=6. $$ - **Direction of motion:** - The object is moving **to the right** when the velocity is positive: $$ 2t-12>0 \quad \Longrightarrow \quad t>6. $$ - The object is moving **to the left** when the velocity is negative: $$ 2t-12<0 \quad \Longrightarrow \quad t<6. $$ Graph the line $$v(t)=2t-12$$ over the domain $$0\le t\le 13$$, marking $$t=6$$ as the stationary point, with the portion for $$t>6$$ above the $$t$$-axis (moving right) and the portion for $$t<6$$ below the $$t$$-axis (moving left). --- **c. Velocity and acceleration at $$t=1$$** - **Velocity at $$t=1$$:** $$ f'(1)=2(1)-12=2-12=-10. $$ So the velocity is $$-10$$ feet per second. - **Acceleration:** The acceleration function is the derivative of the velocity function: $$ f''(t)=\frac{d}{dt}(2t-12)=2. $$ The acceleration is constant $$2$$ feet per second squared. Thus, at $$t=1$$, $$ f''(1)=2. $$ --- **d. Acceleration when the velocity is zero** The velocity is zero at $$t=6$$. Since the acceleration is constant, $$ f''(6)=2. $$ Thus, when the object's velocity is zero, the acceleration is $$2$$ feet per second squared. --- **e. Intervals where the speed is increasing** The speed is the absolute value of the velocity, $$|f'(t)|$$. To determine where the speed is increasing, we analyze the sign of the velocity and the acceleration simultaneously. In general, speed is increasing when the velocity and the acceleration have the same sign. - The acceleration is constant and positive ($$2$$) for all $$t$$. - **When $$t>6$$:** - The velocity $$f'(t)=2t-12$$ is positive. - Since both velocity and acceleration are positive, the speed is increasing. - **When $$t<6$$:** - The velocity is negative. - Here the acceleration is positive while the velocity is negative. With a positive acceleration, the magnitude $$|f'(t)|$$ (which equals $$-f'(t)$$ when $$f'(t)<0$$) is decreasing. Therefore, the speed is increasing on the interval $$ (6,13]. $$ --- **Summary of Answers** - **Stationary:** The object is stationary at $$t=6$$. - **Moving to the right:** The object is moving to the right on the interval $$(6,13]$$. The object is stationary at $$ t = 6 $$ and moving to the right on the interval $$ (6, 13] $$.

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