Question

a body starts from rest and accelerates with acceleration a and distance is S after it moves constant speed in time t
. and it deaccelerates with a/2.and total distance is 15S
Find S

Answer 1

body starts from rest so ,
initial velocity, u=0, let final velocity be v
acceleration= a
distance covered= S
we know,
v^2=u^2+2aS
or, v^2=0+2aS
or, v= sqrt(2aS)
Then it moves with constant speed for time t.
let distance covered be S1
speed=distance/time
so, distance=speed*time
S1=sqrt(2aS)*t
Then it retards with a/2 till it comes to rest
let the distance covered be S2
again applying, v^2=u^2+2aS
0=2aS-aS2
S2= 2S.
total distance covered is 15S
by the problem,
S+S1+S2=15S
or, S+ t*sqrt2aS+ 2S= 15S
or, t*sqrt2aS= 12S
or,t^2*(2aS)=144S^2
or, S=(a*t^2)/72