Mechanics Question need help pls (1 Viewer)

[=)(=]

The acceleration of a particle moving in a straight line is given by a=k(1-v^2),k>0 where v is its velocity at any time t. Initially the particle is at the origin and at rest.
a) Find an expression for the velocity in terms of t and hence the velocity as t->infinity
b) Find an expression for the position of the particle in terms of velocity.

 

[ExtremelyBoredUser]

So pretty much just substitute acceleration with dv/dt.

Rearrange to get t alone:

dt = 1/(k)(1-v)^2 dv

Then integrate both sides;

t = 1/k * int(1/1-v^2) dv

Use partial fractions to find the answer:

t= 1/k * int(1/(1-v^2)) dv

t = 1/k * int(1/(1+v)(1-v) dv

partial fractions decomposition;

1/(1+v)(1-v) = a/(1+v) + b(1-v)

1 = a(1-v) + b(1+v)

1 = a - av + b - bv
1 = v(b-a) + a + b

a + b = 1,
b - a = 0,
2b = 1 therefore b = 1/2 and a = 1/2 since b = a

t =1/k * int[ 1/2(1+v) + 1/2(1-v) dv]

t = 1/k * [ln(1+v) +ln(1-v)/2]

t = 1/k * ln(1-v^2)/(2) +c

and u can resub the original limits to find that C = 0

2kt = ln(1-v^2)
e^2kt = 1-v^2

v^2 = 1 - e^2kt

v = sqrt(1-e^2kt)

so as t -> inf, e^2kt = 0;

v = sqrt(1) = 1

hence v = 1 as t goes to inf

b)

dv/dx * v =k(1-v^2)

dx/(vdv) = 1/(k)(1-v^2)
dx = 1/k * v/1-v^2 dv

x = -1/2k int(-2v/1-v^2) dv
x = -1/2k ln(1-v^2) +c

c = 0 using original boundaries, feel free to check

x = -1/2k ln(1-v^2)

i would type this out on latex but its really late

p.s I probably made a mistake or smthn so please check.

 

[CM_Tutor]

I agree with part (a) up to

t =1/k * int[ 1/2(1+v) + 1/2(1-v) dv]

but there is an error in the integration and in the subsequent reasoning that

v = sqrt(1-e^2kt)

so as t -> inf, e^2kt = 0;

v = sqrt(1) = 1

as as .

Starting from where I agree, we have:

​

The particle starts from the origin, where it is stationary but has a positive acceleration. Thus, it moves in a positive direction with a velocity that increases from 0 and which asymptotically approaches 1 distance unit per time unit from below - and thus is consistent with the above working. The situation would be different if the velocity ever exceeded 1 (such as starting at 2) but that would require a different set of working (like the different paths through a slope field).

For part (b), I agree that the result is and thus, as and thus . It follows that and that the particle continues moving in the positive direction without bound for all time but with its velocity approaching but never reaching 1.

 

[ExtremelyBoredUser]

I agree with part (a) up to

but there is an error in the integration and in the subsequent reasoning that

as as .

Starting from where I agree, we have:

​

The particle starts from the origin, where it is stationary but has a positive acceleration. Thus, it moves in a positive direction with a velocity that increases from 0 and which asymptotically approaches 1 distance unit per time unit from below - and thus is consistent with the above working. The situation would be different if the velocity ever exceeded 1 (such as starting at 2) but that would require a different set of working (like the different paths through a slope field).

For part (b), I agree that the result is and thus, as and thus . It follows that and that the particle continues moving in the positive direction without bound for all time but with its velocity approaching but never reaching 1.

Ah okay, thanks for pointing that out, the mistake I made was that I forgot to put a -ve sign in front of the ln(1-v) as the integral would've been -1/2 * int[(-1/1-v) dv] = - ln(1-v). I did it in a hurry so I'll try not to skip steps next time. Appreciate it!