A point moves in a straight line so that it's distance x from a fixed point at any time t is equal to ktn , where k is any constant. If v be the velocity and f be the acceleration at any time, prove that v2 = nfx/(n-1).

Dear student,
Given that x = ktn    -------(1)
velocity, v = dx/dt = nktn-1   -------(2)
acceleration, f = d2x/dt2 = dv/dt = n(n-1)k tn-2   -----(3)

From (2) and (1) we can write, v = x(n/t)
on squaring   v2 = x2 (n/t)2 = xn2 / t2  ------(4)
from (3) we have f = n(n-1)k tn / t2,
or, t2 =  n(n-1)k tn / f     ------(5)
put t2 from (5) into (4), we get
v2 =  f xn2 / n(n-1)k tn 
v2 =  f xn2 / n(n-1) x
v2 =  f x n / (n-1)

Regards

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