According to the question
a=vdxdv=v(dxdβx−2n)
=(−2nβx−2n−1)βx−2n
=−2nβ2x−4n−1
NEET UG 2015 — Physics Mechanics
A particle undergoes a one-dimensional motion such that its velocity varies according to v(x)=βx−2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by
Held on 30 Apr 2015 · Verified 9 Jul 2026.
−2nβ2x−2n−1
−2nβ2x−4n−1
−2nβ2x−2n+1
−2nβ2x−4n+1
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