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. A particle moves along the x axis according to the equation x=2.00+3.00t -1.00t^2, where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. Hint x = x_0 + v_0t + 1/2at^2,v= (dx)/(dt) =v_0 + at Answer (a) 2.00 m (b) -3.00m//s (c) -2.00 m//s^s Show Steps Givens: x=2.00+3.00t -1.00t^2 t = 3.00s (a) at t=3.00s x = 2.00 + 3.00(3.00) - 1.00(3.00)^2 = color(red)(2.00m) (b) v = (dx)/(dt) = 3.00 -2.00t at t = 3.00s v = 3.00 - 2.00(3.00) = color(red)(-3.00 m//s) (c) a = (dv)/(dt) =-2.00m//s^2 at t=3 a = color(red)(-2.00m//s^2)