A particle moves with an initial velocity u and retardation αv, where v is the velocity at any time t...? ?

Question

A particle moves with an initial velocity u and retardation αv, where v is the velocity at any time t, then:

A) The particle will cover a total distance u/α

B) the particle will come to rest after a time 1/α

C)The particle will continue to move for a very long time

D)The velocity of the particle will become u/2 after a time 1/α

More than one options may be correct.

Answer 1

A) and C) are correct.

Explanation:

Equation of motion:

a = dv/dt = - α∙v

or with v=ds/dt

d²s/dt² + α∙ds/dt = 0

This is a linear differential with constant coefficients, which has the characteristiceristic polynomial

z² + α∙z = 0

with the roots

z₁= 0 and z₂ = -α

Hence:

s = C₁∙exp(z₁∙t) + C₂∙exp(z₂∙t) = C₁ + C₂∙exp(-α∙t)

and

v = ds/dt = -α∙C₂∙exp(-α∙t)

Apply initial condition to find constants

(i)

t=0 → v=u

<=>

u =  -α∙C₂∙exp(-α∙0)

<=>

C₂ = -u/α

(ii)

t=0 → s=0

0 = C₁  - (u/α)∙exp(-α∙0)

<=>

C₁ = (u/α)

Therefore

s = (u/α)∙(1 - exp(-α∙t))

and

v = u∙exp(-α∙t)

The exponential term diminishes for increasing times and vanishes as t approaches infinity. But for finite times is always (a bit) greater. So the particle will move for an infinitely long time. Actually at some time the the motion becomes unobservable cause exponential term gets too small. However answer C is correct.

At the time 1/α it is still moving with

v = u∙exp(-1) = u/e ≈ 0.368∙u

So neither answer B) nor answer C) are correct.

For long times particle approaches a finite position:

limt→∞ {s} = limt→∞ {(u/α)∙(1 - exp(-α∙t))}

= (u/α)∙(1  -  limt→∞ {exp(-α∙t))}

= (u/α)

Therefore answer A) is correct.