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. An object moves with constant acceleration 4.00 m//s^2 and over a time interval reaches a final velocity of 12.0 m//s. (a) If its initial velocity is 6.00 m//s, what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its initial velocity is -6.00 m//s, what is its displacement during the time interval? (d) What is the total distance it travels during the interval in part (c) ? Hint v_f^2 = v_i^2 +2a∆x Answer (a)13.5m (b)13.5m (c)13.5m (d)22.5m Show Steps Givens: a=4.00 m//s^2 v_f = 12.0 m//s (a) Given:v_i = 6.00 m//s v_f^2 = v_i^2 +2a∆x ∆x = (v_f^2 - v_i^2)/(2a) ∆x = ((12.0 m//s)^2 - (6.00 m//s)^2)/(2(4.00 m//s^2)) = color(red)(13.5m) (b) Since the acceleration and velocity of the object are in the same direction. distance equals displacement. d= color(red)(13.5m) (c) Given:v_i = -6.00 m//s v_f^2 = v_i^2 +2a∆x ∆x = (v_f^2 - v_i^2)/(2a) ∆x = ((12.0 m//s)^2 - (-6.00 m//s)^2)/(2(4.00 m//s^2)) = color(red)(13.5m) (d) Since the acceleration and velocity of the object are not in the same direction. distance equals displacement of v= -6 m//s to v=0 plus displacement of v=0 to v=12.0 m//s. Fist part: v_i = -6.00 m//s v_f = 0 ∆x_1 = ((0 m//s)^2 - (-6.00 m//s)^2)/(2(4.00 m//s^2)) = -4.50 m Second part: v_i = 0 m//s v_f = 12.0 m//s ∆x_1 = ((12.0 m//s)^2 - (0 m//s)^2)/(2(4.00 m//s^2)) = 18.0 m distance: d = |∆x_1| + |∆x_2| = |-4.50 m| + |18.0 m| = color(red)(22.5m)