Question

A particle executes the motion described by 

x(t) = x₀(1−e^-yt);  t ≥ 0, x₀ > 0. 

(a) Where does the particles start and with what velocity?

(b) Find maximum and minimum values of x(t), v(t), a(t). Show that x(t) and a(t) increase with time and v(t) decrease with time.

Answer 1

As x(t) = x₀(1-e^-yt)

(a) For velocity,  v(t) = \(\frac{dx(t)}{dt}=+x_0γe^{-yt}\)

For acceleration,

\(a(t) = \frac{dv(t)}{dt}=-x_0\gamma^2e^{\sqcup\gamma t}\)

(b) When t = 0; x(t) = x₀(1−e⁻⁰) = x₀(1−1) = 0

v(t = 0) = x₀γe⁻⁰ = x₀γ(1) = γx₀

(c) (i) x(t) is maximum, when-

t = ∞, x(t) = x₀

x(t) is minimum, when t = 0, x(t) = 0

(ii) v(t) is maximum, when t=0, v(0) = x₀γ

v(t) is minimum, when t = ∞, v(∞) = 0

(iii) a(t) is maximum, when t = ∞, a(∞) = 0

a(t) is minimum, when t = 0, a(0) = -x₀γ