Moving object in the positive direction of the x-axis with a relationship, as the following, x(t)=8+2t+3t^3 Where the distance is measured in meters and the time in second. A. find the instantaneous velocity of the object after two seconds? B. find the instantaneous acceleration of the object after two seconds? C. find the distance of the object after two seconds?

### Given:<br />The position of the object is given by: <br />\[<br />x(t) = 8 + 2t + 3t^3<br />\] <br />where \(x(t)\) is in meters and \(t\) is in seconds.<br /><br />---<br /><br />### A. **Find the instantaneous velocity after 2 seconds** <br />The instantaneous velocity is the first derivative of the position function with respect to time (\(v(t) = \frac{dx(t)}{dt}\)): <br />\[<br />v(t) = \frac{d}{dt}[8 + 2t + 3t^3] = 0 + 2 + 9t^2<br />\] <br />Substitute \(t = 2\): <br />\[<br />v(2) = 2 + 9(2)^2 = 2 + 9(4) = 2 + 36 = 38 \, \text{m/s}<br />\] <br /><br />**Answer:** The instantaneous velocity after 2 seconds is **38 m/s**.<br /><br />---<br /><br />### B. **Find the instantaneous acceleration after 2 seconds** <br />The instantaneous acceleration is the first derivative of the velocity function with respect to time (\(a(t) = \frac{dv(t)}{dt}\)), or equivalently, the second derivative of the position function: <br />\[<br />a(t) = \frac{d}{dt}[2 + 9t^2] = 0 + 18t<br />\] <br />Substitute \(t = 2\): <br />\[<br />a(2) = 18(2) = 36 \, \text{m/s}^2<br />\] <br /><br />**Answer:** The instantaneous acceleration after 2 seconds is **36 m/s²**.<br /><br />---<br /><br />### C. **Find the distance of the object after 2 seconds** <br />The distance is simply the position function evaluated at \(t = 2\): <br />\[<br />x(2) = 8 + 2(2) + 3(2)^3<br />\] <br />Simplify: <br />\[<br />x(2) = 8 + 4 + 3(8) = 8 + 4 + 24 = 36 \, \text{m}<br />\] <br /><br />**Answer:** The distance of the object after 2 seconds is **36 m**.