Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 1. An object moves along the x axis according to the equation x = 3.00t^2 - 2.00t + 3.00, where x is in meters and t is in seconds. Determine (a) the average speed between t = 2.00 s and t = 3.00 s, (b) the instantaneous speed at t = 2.00 s and at t = 3.00 s, (c) the average acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s. (e) At what time is the object at rest? Hint x = x_0 + v_0t + 1/2at^2,v= (dx)/(dt) =v_0 + at,a=(dv)/dt Answer (a) overline(v) = 13.0 m//s (b) v_(2.00s) = 10.0 m//s v_(3.00s) = 16.0 m//s (c) overline(a) = 6.00 m//s^2 (d) a_(2.00s) = 6.00 m//s^2 a_(3.00s) = 6.00 m//s^2 (e) t=0.333s Show Steps (a) x_(2.00) = 3.00(2.00)^2-2.00(2.00) +3.00 =11.0m x_(2.00) = 3.00(3.00)^2-2.00(3.00) +3.00 =24.0m overline(v) = (24.0 m - 11.0 m)/(3.00s-2.00s) =color(red)(13.0 m//s) (b) v = (dx)/(dt) = 6.00t -2.00 v_(2.00) = 6.00(2.00) -2.00 = color(red)(10.0 m//s) v_(3.00) = 6.00(3.00) -2.00 = color(red)(16.0 m//s) (c) a = (dv)/(dt) = 6.00 v_(2.00) = color(red)(6.00 m//s^2) v_(3.00) = color(red)(6.00 m//s^2) (e) when object is at rest v=0 m//s we get: 6.00t - 2.00 = 0 t = 2.00/6.00 = color(red)(0.333 s)