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 $$ s0 $$ corresponding to 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? $$ f(t)=t^{2}-12 t ; 0 \leq t \leq 13 $$ $$ t=6 $$ (Simplify your answer. Use a comma to separate answers as needed.) When is the object moving to the right? The object is moving to the right on $$ (6,13] $$. (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.) When is the object moving to the left? The object is moving to the left 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 $$ s>0 $$ corresponding to 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? $$ f(t)=t^{2}-12 t ; 0 \leq t \leq 13 $$ $$ t=6 $$ (Simplify your answer. Use a comma to separate answers as needed.) When is the object moving to the right? The object is moving to the right on $$ (6,13] $$. (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.) When is the object moving to the left? The object is moving to the left on (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.)

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**a. Graph the Position Function** The position function is $$ f(t)=t^2-12t,\quad 0\le t\le 13. $$ This is a parabola opening upward. Its vertex occurs at $$ t=-\frac{b}{2a}=-\frac{-12}{2\cdot 1}=6. $$ At $$ t=6 $$, the position is $$ f(6)=6^2-12\cdot6=36-72=-36. $$ Thus the vertex is at $$ (6,-36) $$. The graph is a U-shaped parabola with endpoints at $$ f(0)=0^2-12\cdot0=0 $$ and $$ f(13)=13^2-12\cdot13=169-156=13. $$ --- **b. Find and Graph the Velocity Function; Determine Motion Directions** The velocity is the derivative of the position: $$ v(t)=f'(t)=\frac{d}{dt}\left(t^2-12t\right)=2t-12. $$ 1. **Stationary:** The object is stationary when $$ v(t)=0 $$: $$ 2t-12=0\quad\Longrightarrow\quad t=6. $$ 2. **Moving to the Right:** The object moves to the right when $$ v(t)>0 $$: $$ 2t-12>0\quad\Longrightarrow\quad t>6. $$ Since $$ t $$ is in $$ [0,13] $$, the object is moving to the right on the interval $$ (6,13] $$. 3. **Moving to the Left:** The object moves to the left when $$ v(t)<0 $$: $$ 2t-12<0\quad\Longrightarrow\quad t<6. $$ Thus, it is moving to the left on $$ [0,6) $$. The graph of the velocity function $$ v(t)=2t-12 $$ is a straight line crossing the $$ t $$-axis at $$ t=6 $$. --- **c. Velocity and Acceleration at $$ t=1 $$** - The velocity at $$ t=1 $$ is: $$ v(1)=2(1)-12=2-12=-10. $$ - The acceleration is the derivative of the velocity: $$ a(t)=v'(t)=\frac{d}{dt}\left(2t-12\right)=2. $$ Hence, at $$ t=1 $$: $$ a(1)=2. $$ --- **d. Acceleration When the Velocity is Zero** The velocity is zero at $$ t=6 $$; therefore, the acceleration at that time is: $$ a(6)=2. $$ --- **e. Intervals on Which the Speed is Increasing** The speed is the absolute value of the velocity, $$ |v(t)| $$. Its increasing behavior depends on both the sign of the velocity and the acceleration. Since the acceleration is constant and positive ($$ a(t)=2>0 $$ for all $$ t $$): - When $$ v(t)>0 $$ (i.e., the object is moving to the right on $$ (6,13] $$), the positive acceleration increases the speed. - When $$ v(t)<0 $$ (i.e., the object is moving to the left on $$ [0,6) $$), the positive acceleration works against the negative velocity, thereby decreasing the magnitude of the velocity. Thus, the speed is increasing on the interval $$ (6,13]. $$ --- **Answer to the Specific Questions** - **When is the object moving to the right?** $$ (6,13] $$. - **When is the object moving to the left?** $$ [0,6) $$. **a. Graph the Position Function** The position function is a parabola opening upward with its vertex at $$ t=6 $$ and $$ s=-36 $$. It starts at $$ (0,0) $$ and ends at $$ (13,13) $$. --- **b. Find and Graph the Velocity Function; Determine Motion Directions** - **Velocity Function:** $$ v(t) = 2t - 12 $$ - **Stationary:** At $$ t=6 $$ - **Moving to the Right:** $$ t > 6 $$ (on $$ (6,13] $$) - **Moving to the Left:** $$ t < 6 $$ (on $$ [0,6) $$) --- **c. Velocity and Acceleration at $$ t=1 $$** - **Velocity:** $$ v(1) = -10 $$ - **Acceleration:** $$ a(1) = 2 $$ --- **d. Acceleration When the Velocity is Zero** - **Acceleration at $$ t=6 $$:** $$ a(6) = 2 $$ --- **e. Intervals on Which the Speed is Increasing** - **Speed Increasing:** $$ (6,13] $$ --- **Answers to Specific Questions** - **When is the object moving to the right?** $$ (6,13] $$ - **When is the object moving to the left?** $$ [0,6) $$

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