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? \( f(t)=t^{2}-12 t ; 0 \leq t \leq 13 \) 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 \( [0,6) \). (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.) c. The velocity of the object at \( t=1 \) is \( \square \square \) (Simplify your answer.)

**a. Graph of the Position Function** The position function is \[ f(t)=t^2-12t \] which is a quadratic function (a parabola) opening upward. Its vertex can be found by computing \[ t_{\text{vertex}}=-\frac{b}{2a}=-\frac{-12}{2\cdot1} =6. \] The corresponding position is \[ s(6)=6^2-12(6)=36-72=-36. \] Thus, the vertex is at \((6,-36)\). The domain is \([0,13]\). On a graph, the parabola starts at \( t=0 \) with \[ s(0)=0^2-12\cdot0=0, \] dips to a minimum of \(-36\) at \( t=6 \), and then increases as \( t \) approaches 13. --- **b. Velocity Function and Motion Direction** The velocity function is the derivative of the position function. Thus, \[ v(t)=f'(t)=\frac{d}{dt}(t^2-12t)=2t-12. \] - **Stationary:** The object is stationary when \( v(t)=0 \). Solve \[ 2t-12=0 \quad \Longrightarrow \quad 2t=12 \quad \Longrightarrow \quad t=6. \] - **Moving to the Right:** The object moves to the right when \( v(t)>0 \). \[ 2t-12>0 \quad \Longrightarrow \quad t>6. \] Hence, the object moves to the right on the interval \((6,13]\). - **Moving to the Left:** The object moves to the left when \( v(t)<0 \). \[ 2t-12<0 \quad \Longrightarrow \quad t<6. \] Thus, the object moves to the left on the interval \([0,6)\). A graph of \( v(t)=2t-12 \) is a straight line with a zero at \( t=6 \) and a positive slope of 2. --- **c. Velocity and Acceleration at \( t=1 \)** - **Velocity at \( t=1 \):** \[ v(1)=2(1)-12=2-12=-10. \] - **Acceleration:** The acceleration function is the derivative of the velocity function. \[ a(t)=v'(t)=\frac{d}{dt}(2t-12)=2. \] Therefore, at \( t=1 \), \[ a(1)=2. \] --- **d. Acceleration When Velocity is Zero** The velocity is zero at \( t=6 \). Since the acceleration is constant at \[ a(t)=2, \] the acceleration when the velocity is zero is \[ a(6)=2. \] --- **e. Intervals Where the Speed is Increasing** The speed is given by the absolute value of velocity: \[ \text{Speed}=|v(t)|=|2t-12|. \] To determine when the speed is increasing, we analyze the relation between velocity and acceleration. In general, the speed increases when the acceleration and velocity have the same sign. - For \( t>6 \): \( v(t)=2t-12>0 \) and \( a(t)=2>0 \). Hence, the speed is increasing on \((6,13]\). - For \( t<6 \): \( v(t)=2t-12<0 \) (moving left) while \( a(t)=2>0 \). In this case, the object is slowing down (the magnitude of velocity is decreasing) so the speed is decreasing. Thus, the speed is increasing only on the interval \((6,13]\). --- **Summary of Answers** - The object is moving to the right on \((6,13]\). - The object is moving to the left on \([0,6)\). - **(c)** The velocity of the object at \( t=1 \) is \[ -10. \]