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 = x2 n2 / 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 x2 n2 / n(n-1)k tn
v2 = f x2 n2 / n(n-1) x
v2 = f x n / (n-1)
Regards
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 = x2 n2 / 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 x2 n2 / n(n-1)k tn
v2 = f x2 n2 / n(n-1) x
v2 = f x n / (n-1)
Regards