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; } 4. A particle moves along the x axis. Its position is given by the equation x=2+3t -4t^2,with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0. Hint v =(dx)/(dt) Answer (a)2.56 m (b)-3m/s Show Steps Given: x = 2 + 3t - 4t^2 (a) from given equation we get: v = (dx)/(dt) = 3 - 8t when v = 0 it changes direction, so  0 = 3-8t => t =3/8 s x = 2 + 3(3/8) -4(3/8)^2 = color(red)(2.56 m) (b) when t=0 x = 2 m 2 = 2 + 3t - 4t^2 t(3-4t) = 0 t= 0 or t = 3/4 s Whent = 3/4 s it returns to the position is had at t = 0 v = 3 - 8t =>v = 3-8(3/4) = color(red)(-3m//s)