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? [? ] and [ ] Enter the smallest first.

To find the times when the object is at rest, we need to find the values of \( t \) for which the velocity \( v(t) \) is equal to zero. Given: \[ v(t) = t^{2} - 14t + 45 \] To find the times when the object is at rest, we set \( v(t) = 0 \) and solve for \( t \). \[ t^{2} - 14t + 45 = 0 \] Now, we can solve this quadratic equation to find the values of \( t \) when the object is at rest. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(t^{2}-14t+45=0\) - step1: Factor the expression: \(\left(t-9\right)\left(t-5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&t-9=0\\&t-5=0\end{align}\) - step3: Solve the equation: \(\begin{align}&t=9\\&t=5\end{align}\) - step4: Rewrite: \(t_{1}=5,t_{2}=9\) The object is at rest at times \( t = 5 \) and \( t = 9 \). Therefore, the two times the object is at rest are 5 and 9.