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 $$ t=9 $$ ? [ ? ]

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We first compute the position function $$ x(t) $$ by integrating the velocity function $$ v(t) $$.

1. The velocity is given by
   $$
   v(t) = t^2 - 14t + 45.
   $$

2. The position is obtained by integrating $$ v(t) $$ from $$ 0 $$ to $$ t $$ (with the initial condition $$ x(0) = 0 $$):
   $$
   x(t) = \int_0^t \left(s^2 - 14s + 45\right) \, ds.
   $$

3. Integrate term-by-term:
   $$
   \int s^2 \, ds = \frac{s^3}{3}, \quad \int -14s \, ds = -7s^2, \quad \int 45 \, ds = 45s.
   $$
   Hence, the position function is:
   $$
   x(t) = \frac{t^3}{3} - 7t^2 + 45t.
   $$

4. To find the $$ x $$-coordinate at $$ t = 9 $$, substitute $$ t = 9 $$ into the position function:
   $$
   x(9) = \frac{9^3}{3} - 7(9)^2 + 45(9).
   $$

5. Calculate each term:
   $$
   \frac{9^3}{3} = \frac{729}{3} = 243,
   $$
   $$
   7(9)^2 = 7 \times 81 = 567,
   $$
   $$
   45(9) = 405.
   $$

6. Combine the results:
   $$
   x(9) = 243 - 567 + 405 = 81.
   $$

Thus, the $$ x $$-coordinate of the object when $$ t = 9 $$ is $$ 81 $$.
At $$ t = 9 $$, the $$ x $$-coordinate is 81.

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 \( t=9 \) ? [ ? ]

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