A car covers 8m in the first second and 12m in the second second. Find a) Initial Velocity
b) Acceleration
c) Distance covered in the third second​

a) u = 6 m/s

b) a = 4 m/s²

c) d(3) = 16 m

Explanation:

equation for the first second

distance will be the average velocity times the time of travel

8 = ½(u + (u + at))t    t is one second, so reduces to

8 = u + ½a

velocity at the end of the first second is

v = u + at = u + a

position equation for the second period is

12 = ½((u + a) + (u + a + at))t   t is one second so reduces to

12 = u + 3a/2

subtracting the first position equation from the second

12 - 8 = u + 3a/2 - (u + ½a)

a = 4 m/s²

8 = u + ½4

u = 6 m/s

in the third second

d = 6(3) + ½(4)(3²) - 8 - 12

d = 16 m

answer; the correct answer is technician a; ///100%good luck

the ima (higher than 1m) will be greater than the ima with the height of 1m.