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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To graph the position function \( f(t) = t^2 - 12t \), you can find the vertex by completing the square or using the vertex formula. The vertex occurs at \( t = 6 \), and evaluating \( f(6) \) gives you the maximum or minimum position. The graph will open upwards (since the coefficient of \( t^2 \) is positive) and will cross the time axis (where the object is at position 0) at points \( t = 0 \) and \( t = 12 \). For the velocity function \( v(t) \), you take the derivative: \[ v(t) = f'(t) = 2t - 12. \] The object is stationary when \( v(t) = 0 \), which happens when \( t = 6 \). To determine when the object is moving right, set \( v(t) > 0 \): \( 2t - 12 > 0 \) leads to \( t > 6 \). So, the object moves to the right for the interval \( (6, 13] \) (since \( t \) must be less than or equal to 13). To find the acceleration \( a(t) \), take the derivative of the velocity: \[ a(t) = v'(t) = 2. \] This shows that the acceleration is constant at 2 feet per second squared at all times, including at \( t = 1 \) where \( a(1) = 2 \) and \( v(1) = 2 \times 1 - 12 = -10 \) ft/s (indicating the object moves left). When the velocity is zero at \( t = 6 \), since the acceleration remains constant at 2, the object is accelerating positively, meaning it begins to change from leftward movement to rightward. The speed is increasing when the object has further positive velocity which is true for \( t > 6 \), so the speed is increasing on the interval \( (6, 13] \). Overall, your answers: The object is stationary at \( t = 6 \). The object is moving to the right on \( (6, 13] \).
