devarshi-dt-logo

Question:

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βxⁿ, 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

ⁿβ²eⁿ⁺¹

ⁿβ²xⁿ⁺¹

β²xⁿ⁺¹

ⁿβ²xⁿ⁺¹

Solution:

v(x) = βxⁿ
a(x) = dv/dt = (dv/dx)(dx/dt) = v(dv/dx)
dv/dx = nβxⁿ⁻¹
v(dv/dx) = βxⁿ(nβxⁿ⁻¹) = nβ²x²ⁿ⁻¹
If n=1, a(x) = β²x
If n=2, a(x) = 2β²x³
Therefore, the acceleration of the particle as a function of x is given by β²xⁿ⁺¹.