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; } 3. The position of a particle moving along the x axis varies in time according to the expression x = 3t^2, where x is in meters and t is in seconds. Evaluate its position (a) at t = 3.00 s and (b) at 3.00 s + ∆t. (c) Evaluate the limit of ∆x//∆t as ∆t approaches zero to find the velocity at t = 3.00 s. Hint f(x) = ax^2 => f'(x) = 2ax Answer (a) 27.0 m (b) 27.0 + 18.0(∆t) + 3.00(∆t)^2 m (c) 18.0 m//s Show Steps (a) When t = 3.00 s x = 3(3.00)^2 = color(red)(27.0 m) (b) When t = 3.00s + ∆t x = 3(3.00 + (∆t))^2 = 3(9.00 + 6.00(∆t) +(∆t)^2) x = color(red)(27.0 + 18.0(∆t) + 3(∆t)^2 m) (c) Solve by Algebra: Here we set x form (a) as x_i set x form (b) as x_f ∆x = x_f - x_i => ∆x = 27.0 + 18.0(∆t) + 3(∆t)^2 - 27.0 => ∆x = 18.0(∆t) + 3(∆t)^2  (∆x)/(∆t) = (18.0(∆t) + 3(∆t)^2)/(∆t) =18.0 +3∆t limit of (∆x)/(∆t) as ∆t approaches zero write as: lim_(∆t->0)(18.0 +3∆t) when ∆t approaches to 0 3∆t = 0  lim_(∆t->0)(18.0 +3∆t) = 18.0 m//s Solve by calculus: x = 3t^2 => v = (∆x)/(∆t) = 2•3t =6t at t = 3.00s v = 6(3.00) m//s = color(red)(18.0 m//s)