A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = x-2n, where 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
- -2 2x-2n+1
- -2n 2e-4n+1
- -2n 2x-2n-1
- -2n 2x-4n-1
Answer
(4) -2n 2x-4n-1
Solution
a = v dvdx
dvdx = (-2n)x-2n-1
⇒ a = x-2n (-2n x-2n-1)
= -2n 2x-4n-1
Answer: (4)