A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx⁻²ⁿ, where b and n are constants and x is the position of the particle. The acceleration of the particle as d function of x is given by
Options
(a) -2nb²x⁻⁴ⁿ⁻¹
(b) -2nb²x⁻²ⁿ⁺¹
(c) -2nb²e⁻⁴ⁿ⁺¹
(d) -2nb²x⁻²ⁿ⁻¹
Correct Answer:
-2nb²x⁻⁴ⁿ⁻¹
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.