# Integral of speed as a function of displacement

by Paul Czerner
Tags: displacement, function, integral, speed
 Homework Sci Advisor HW Helper Thanks P: 9,651 You have dx/dt = f(x)? Turn that into dt = dx/f(x) and integrate, if you can: t = $\int$(1/f(x)).dx However, you mentioned path integrals, so maybe there's more than one space dimension here? Please provide the actual equation.
 Homework Sci Advisor HW Helper Thanks P: 9,651 Integral of speed as a function of displacement As I understand it, you have speed v = v(x), so t = ∫(1/v(x)).dx If you want to know the position x1 reached at time t1 then you have to solve t1 = $∫^{x1}_{0}$(1/v(x)).dx E.g. suppose v(x) = a.x + b t1 = [ln(x+b/a)/a$]^{x1}_{0}$ = (ln(x1+b/a) - ln(b/a)]/a a.x1 = b.exp(a.t1) - b If that's not what you're trying to do, please pick a specific v=v(x) so that we an discuss it more clearly.
 Quote by haruspex As I understand it, you have speed v = v(x), so t = ∫(1/v(x)).dx If you want to know the position x1 reached at time t1 then you have to solve t1 = $∫^{x1}_{0}$(1/v(x)).dx E.g. suppose v(x) = a.x + b t1 = [ln(x+b/a)/a$]^{x1}_{0}$ = (ln(x1+b/a) - ln(b/a)]/a a.x1 = b.exp(a.t1) - b If that's not what you're trying to do, please pick a specific v=v(x) so that we an discuss it more clearly.