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.

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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.
The object is at rest at times 5 and 9.

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.

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