Question

a particle moves in straight line with acceleration described by a=mx-v²/x. if the initial velocity and displacement are v and zero and at any time t velocity and displacement are 0 and x. the value of constant m is​

Answer 1

Answer:

m = 3v² / 2x²

Explanation:

At, T=0..

initial velocity v₀ = v and displacement s₀ = 0

At, T=t..

initial velocity v = 0 and displacement s = x

Now according to the law of kinematics.

v² = v₀² + 2×a×(s-s₀)

►a = $\frac{(V^2 - V_o ^2)} {2(S- S_o)}$ = $\frac{0 - v^2} {2x}$ = v² / 2x

So, mx - v²/x = v² / 2x

►mx= 3v² / 2x

► m = 3v² / 2x²