Question

A particle executes the motion described by x(t) = x₀(1-(e^-γt))
t≥0, x₀>0.
(a) Where does the particle start and with what velocity?
(b) Find maximum and minimum values of x(t), v(t) and a(t).
Show that x(t) and a(t) increase with time and v(t) decreases with time.​

Answer 1

Answer:

Explanation:

Equation of motion of a particle

x(t) = x0(1−e−yt); t≥0, x0>0−−−(1)

(a) particle starts from x(0) =0, at t=0

dx(t)/dt = x0(0−y×e^−yt)

             =x0×y×e^−yt

                 =− y×[x(t)−x0].−−−−(2)

particle starts with velocity v= dx(0)/dt=x0×y.

The acceleration of particle a= dv/dt=d2x(t)dt2=−x0(y^2×e^−yt).

a=−x0 y^2 e^−yt

  = y^2[x(t)−x0]−−−(3)

From eqns (1), (2) and (3) it is known that

(b) maximum values of x(t),

v(t) and a(t) are xo, −x0 y  and 0.    

the minimum values of  x(t), v(t) and a(t) are 0, 0 and  −x0 y^2.

from eqns (1), (2)  and (3) it is known that x(t) and a(t) increase with time whereas v(t)decreases with time by noting their signs.