When the object starts to move in a straight line with uniform acceleration from its rest, it is known that the difference between the displacement in the 4th and 2nd seconds is 12M, and then (1) the acceleration of the object moving and (2) the average velocity in the 4th second can be obtained A car is moving in a straight line with uniform acceleration. It takes 5 seconds to pass two trees 80 meters apart on the roadside. The speed when it passes the first tree is 8 MGS. Find the speed when it passes the second tree I have to do it today,

When the object starts to move in a straight line with uniform acceleration from its rest, it is known that the difference between the displacement in the 4th and 2nd seconds is 12M, and then (1) the acceleration of the object moving and (2) the average velocity in the 4th second can be obtained A car is moving in a straight line with uniform acceleration. It takes 5 seconds to pass two trees 80 meters apart on the roadside. The speed when it passes the first tree is 8 MGS. Find the speed when it passes the second tree I have to do it today,

(1) From the formula & nbsp; xm-xn = (m-n) at & # 178; & nbsp; t is the time interval 1s & nbsp; & nbsp; it is obtained that:
  x4-x2=(4-2)a×1²=12
a=6m/s²
(2) The initial velocity V1 and the final velocity V2 in the fourth second;
v1=at1=6×3=18m/s               v2=at2=6×4=24m/s     
Average velocity = (V1 + V2) / 2 = (18 + 24) / 2 = 21m / S
x=vot+½at²
80=8×5+½a×5²
a=3.2m/s²
v=vo+at=8+3.2×5=24m/s