A body moving with constant acceleration travels the distance 6 m and 16 m respectively in 2 s and 4 s. Calculate the initial velocity and acceleration of the body

Answer 1

let the acceleration be a m/s^2
v = at + c
s = (1/2)a t^2 + ct + k

at the beginning of our observation:
t = 0 , s = 0 , then k = 0

s = (a/2) t^2 + ct

when t = 2
6 = (a/2)(4) + 2c
6 = 2a + 2c
a+c = 3

when t = 4
16 = (a/2)(16) + 4c
16 = 8a + 4c
4 = 2a + c

subtract those last two equations:
c = 2
and then a = 1

the acceleration is 1 m/s^2
when t = 0 (initial)
v = at + c
= 1(0) + 2
= 2

and the initial velocity is 2 m/s

I did this using Calculus. They have formulas for this in
physics, perhaps using them would be shorter and simpler.