The velocity of an object moving along the \( x \)-axis is \( v(t)=t^{2}-14 t+45 \). The initial position is the origin. What are the two times the object is at rest? \[ 5 \text { and } 9 \] What is the \( x \)-coordinate at \( t=9 \) ? 81 What is the acceleration at \( t=9 \) ? [? ]

To find the acceleration at \( t = 9 \), we first need to calculate the acceleration function, which is the derivative of the velocity function \( v(t) = t^{2} - 14t + 45 \). Taking the derivative, we get \( a(t) = v'(t) = 2t - 14 \). Now, substituting \( t = 9 \) into the acceleration function: \[ a(9) = 2(9) - 14 = 18 - 14 = 4. \] So, the acceleration at \( t = 9 \) is \( 4 \, \text{units/s}^{2} \). This showcases how derivatives are not just for finding slopes but also allow us to sift through the motions of objects, revealing secrets like how fast they're speeding up or slowing down at given moments! Additionally, understanding acceleration helps us explore the concepts of force in the real world. Imagine a roller coaster: as it climbs, the acceleration is negative (deceleration), and as it descends, it’s positive. That thrill comes from the interplay between velocity and acceleration!