An object moves with constant acceleration 485 and over a ti

An object moves with constant acceleration 4.85 and over a time interval reaches a final velocity of 11.2 if its original velocity is 5.60 what is its displacement during the time interval what is the distance it travels during this interval? If its initial velocity is -5.60 m/s, what is its displacement during this interval? What is the total distance it travels during the interval in part(c)?

Solution

(a) The acceleration is the change in velocity over an interval of time. Let t be the time interval in which the velocity changes from 5.60 m/s to 11.2 m/s with a constant acceleration of 4.85 m/s² . Then 4.85 = ( 11.20 - 5.60)/ t or,

4.85 = 5.60/t. Therefore, t = 5.60/4.85 = 1.15464 seconds ( approx.on rounding off to 5 decimal places)

Also the average velocity = (5.60 + 11.20)/2 = 16.80/2 = 8.40 m/s. The displacement = average velocity x time = 8.40* 1.15464 = 9.698976 m = 9.70 m ( on rounding off to 2 decimal places).

(b) Since no chane of direction is mentioned, the distance travelled is the same as displacement i. e. 9.70 m

(c) If the initial velocity is - 5.50 m/s, then 4.85 = [ 11.20 - (- 5.60)] / t   or, 4.85 = ( 11.20 + 5.60)/t = 16.80/t . Therefore, t = 16.80/4.85 = 3.4639 sec.The average velocity = (- 5.60 + 11.20 )2 = 5.60/2 = 2.80 m/s. The  displacement = average velocity x time = 2.80*3.4639 = 9.6989 m = 9.70 m ( on rounding off to 2 decimal places).

(d) Since no chane of direction is mentioned, the distance travelled is the same as displacement i. e. 9.70 m