An object moves with constant acceleration 4.70 m/s^2 and over a time interval r
ID: 2219726 • Letter: A
Question
An object moves with constant acceleration 4.70 m/s^2 and over a time interval reaches a final velocity of 11.8 m/s. (a) If its initial velocity is 5.9 m/s, what is its displacement during the time interval? (Please provide numerical answer in meters) (b) What is the distance it travels during this interval? (Please provide numerical answer in meters) (c) If its initial velocity is -5.9 m/s, what is its displacement during the time interval? (Please provide numerical answer in meters) (d) What is the total distance it travels during the interval in part (c)? (Please provide numerical answer in meters)Explanation / Answer
a)v^2 - u^2 = 2as so s = 11.8^2 - 5.9^2 /2*4.7 = 11.1 meters b)as direction is same distance is also same = 11.1 meters c) displacement = same again as first it goes in right directions attains 0 and come to same point with +5.9 m/s so answer = 11.1 meters d) distance = 11.1 + 2(5.9^2/2*4.7) = 18.5 meters if any douts comment i will explain please rate good