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 \( t=9 \) ? [ ? ]

To find the \( x \)-coordinate at \( t=9 \), we first need to find the position function by integrating the velocity function \( v(t) = t^2 - 14t + 45 \). Integrating \( v(t) \) gives us the position function: \[ x(t) = \int (t^2 - 14t + 45) dt = \frac{t^3}{3} - 7t^2 + 45t + C. \] Since the initial position is the origin, \( x(0) = 0 \), we find \( C = 0 \). Now, we calculate \( x(9) \): \[ x(9) = \frac{9^3}{3} - 7(9^2) + 45(9) = \frac{729}{3} - 7(81) + 405 = 243 - 567 + 405 = 81. \] Thus, the \( x \)-coordinate at \( t=9 \) is \( 81 \).