Question

A particle executes the motion described by x(t) = x₀ (1 − e − γt); t ≥ 0, x₀ > 0. Where does the particle start and with what velocity?

Answer 1

Given, `x(t) = x_0 (1 - e^(-γt))`

`v(t) = (dx(t))/(dt) = x_0 γe^(-γt)`

`a(t) = (dv(t))/(dt) = - x_0 γ^2 e^(-γt)`

When t = 0; x(t) = x₀(1 – e^–0) = x₀ (1 – 1) = 0

x(t = 0) = x₀γe^–0 = x₀γ(1) = γx₀