102=vi(6)+1/2 a 36
18=vi(2)+ 1/2 a 4
multiply the second equation by 3, then subtract the second equation from the first.
an object travels in a straight line at a constant acceleration for 6 seconds.in the first 2 seconds it travels 18 metres and in entire 6 seconds it covers 102 metres. find the acceleration.
3 answers
d = Vi t + (1/2) a t^2
18 = Vi (2) + (1/2) a (4)
102 = Vi (6) + (1/2) a (36)
18 = 2 Vi + 2 a
or
9 = Vi + a
102 = 6 Vi + 18 a
or
17 = Vi + 3 a
8 = 2 a
a = 4 m/s^2
and
Vi = 5 m/s
18 = Vi (2) + (1/2) a (4)
102 = Vi (6) + (1/2) a (36)
18 = 2 Vi + 2 a
or
9 = Vi + a
102 = 6 Vi + 18 a
or
17 = Vi + 3 a
8 = 2 a
a = 4 m/s^2
and
Vi = 5 m/s
thnx guyz !!!!!!!!