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Q.

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according tvx=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 :

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a

-2nβ2x-4n-1

b

-2nβ2x-2n+1

c

-2nβ2x-4n+1

d

-2nβ2x-2n-1

answer is A.

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Detailed Solution

v=βx−2nSo, dvdx=−2nβx−2n−1Now a=vdvdx=βx−2n−2nβx−2n−1⇒a=−2nβ2x−4n−1
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