a body moving in constant acceleration describes successive equal distances in time intervals x y and z prove that 3/(x+y+z)=1/x-1/y+1/z

Answer 1

Let the speed at the beginning ox x interval be u, Let d be distance covered in each interval, and a be the acceleration.

distance covered in x interval, d = ux+.5ax²

Or d/x = u +.5ax

Speed in the beginning of y = u + ax

Distance covered in y interval, d = (u+ax)y +.5ay²

Or d/y = u+ ax+ .5 ay

Similarly Speed in the beginning of z interval = u+a(x+y)

So d/z = u +a(x+y) +.5az

d/z= u + a(x+y+.5z)

Now d/x -d/y +d/z = u +.5ax -(u +ax+.5y) + u+ax+ay +.5az = u+.5ax+.5ay+.5az

i.e. d(1/x-1/y+1/z) = u + .5a(x+y+z) --(1)

3d is the distance covered in the interval x+y+z, with starting speed u

So 3d/(x+y+z) =u+.5(x+y+z) ---(2)

Equating (1) and (2), we get d(1/x-1/y+1/z) = 3d/(x+y+z)

Or (1/x-1/y+1/z) = 3/(x+y+z)

Answer 2

Let the speed at the beginning of x interval be u, Let d be distance covered in each interval, and a be the acceleration.

distance covered in x interval, d = ux+.5ax²

Or d/x = u +.5ax

Speed in the beginning of y = u + ax

Distance covered in y interval, d = (u+ax)y +.5ay²

Or d/y = u+ ax+ .5 ay

Similarly Speed in the beginning of z interval = u+a(x+y)

So d/z = u +a(x+y) +.5az

d/z= u + ax+ay+.5az

Now d/x -d/y +d/z = u +.5ax -(u +ax+.5y) + u+ax+ay +.5az = u+.5ax+.5ay+.5az

i.e. d(1/x-1/y+1/z) = u + .5a(x+y+z) --(1)

3d is the distance covered in the interval x+y+z, with starting speed u

So 3d/(x+y+z) =u+.5(x+y+z) ---(2)

Equating (1) and (2), we get d(1/x-1/y+1/z) = 3d/(x+y+z)

Or (1/x-1/y+1/z) = 3/(x+y+z)