# Mechanics (Application of Integration)

#### Muddyakka

##### New member
A particle moving in a straight line experiences a force numerically equal to 2/x2 per unit mass towards the origin. Find the time taken to reach the origin if the particle starts at rest 4 units from the origin.
I have been able to arrive at an equation:
a(x) = 2/x2 (as force per unit mass, I am assuming mass to be 1 unit)
a(x) = dv/dt = dv/dx . dx/dt = v dv/dx (chain rule)
Hence, v dv = (2/x2) dx
v2/2 = (-2/x2) + c
v2 = (-4/x) + C
but at x=4, v=0
Hence, C = 1

My final equation: v2 = (-4/x) +1
But from the looks of it trying to find x(t) and then setting to zero seems too difficult. How else can you solve this? The answer is 2 pi seconds

#### Jomo

##### Elite Member
IF v2/2 = (-2/x2) + c, then how do you then get v2 = (-4/x) + C. That is not correct.

If v dv = (2/x2) dx, then how did you get v2/2 = (-2/x2) + c

#### Subhotosh Khan

##### Super Moderator
Staff member
A particle moving in a straight line experiences a force numerically equal to 2/x2 per unit mass towards the origin. Find the time taken to reach the origin if the particle starts at rest 4 units from the origin.
I have been able to arrive at an equation:
a(x) = 2/x2 (as force per unit mass, I am assuming mass to be 1 unit)
a(x) = dv/dt = dv/dx . dx/dt = v dv/dx (chain rule)
Hence, v dv = (2/x2) dx
v2/2 = (-2/x) + c
v2 = (-4/x) + C
but at x=4, v=0
Hence, C = 1

My final equation: v2 = (-4/x) +1
But from the looks of it trying to find x(t) and then setting to zero seems too difficult. How else can you solve this? The answer is 2 pi seconds
Assuming your calculations above are correct

$$\displaystyle v = \ \pm \ \sqrt{1 \ - \ \frac{4}{x}}$$

substitute:

x = 4 * sec2(u)

dx/du = 4 * 2 * sec2(u) * tan(u)

$$\displaystyle v = dx/dt = \ \pm \ sin(u)$$

Now continue.....

Problems in engineering are generally difficult (- otherwise everybody would become engineer and destroy our club)

Last edited:

#### Jomo

##### Elite Member
Problems in engineering are generally difficult (- otherwise everybody would become engineer and destroy our club)
And you have a PhD in engineering. I am impressed! Now, lets see if you can get an advanced degree in math. The math club is more exclusive!

#### HallsofIvy

##### Elite Member
But doesn't have as much money.

#### Muddyakka

##### New member
IF v2/2 = (-2/x2) + c, then how do you then get v2 = (-4/x) + C. That is not correct.

If v dv = (2/x2) dx, then how did you get v2/2 = (-2/x2) + c
Sorry, that was a transcription error.

Thinking again:
a(x) = -2/x2
vdv/dx = -2/x2
v2/2 = (2/x) + c
v2 = (4/x) + C
sub x=4, v=0
C = -1
Hence v2 = (4/x) -1

#### Subhotosh Khan

##### Super Moderator
Staff member
Sorry, that was a transcription error.

Thinking again:
a(x) = -2/x2
vdv/dx = -2/x2
v2/2 = (2/x) + c
v2 = (4/x) + C
sub x=4, v=0
C = -1
Hence v2 = (4/x) -1
In that case , follow the "process" shown in response #3, but first substitute

x = 4 cos2(Θ)

and continue....

#### Muddyakka

##### New member
Assuming your calculations above are correct

$$\displaystyle v = \ \pm \ \sqrt{1 \ - \ \frac{4}{x}}$$

substitute:

x = 4 * sec2(u)

dx/du = 4 * 2 * sec2(u) * tan(u)

$$\displaystyle v = dx/dt = \ \pm \ sin(u)$$

Now continue.....

Problems in engineering are generally difficult (- otherwise everybody would become engineer and destroy our club)
Thanks for the advice! In hindsight, I should have noticed that a pi should indicate a trig sub but anyways.
I tried x = 4sin2(a)
time taken was a definite integral
x: 0-4
a: 0-pi/2
dx = 8sin(a)cos(a) da
Hence integrate:
8sin2(a) da
Using: cos(2a) = 1 - 2sin2(a)
2sin2(a) = 1 - cos(2a)
Now integrate:
4(1 - cos(2a))
Which gives:
4[a-(sin(2a)/2)] with bounds 0 and pi / 2
Giving 2pi seconds as the final answer.

#### Jomo

##### Elite Member
But doesn't have as much money.
You know better than that. I am sure that you studied math for the beauty of it. Money nor application never came to mind!
Besides, Khan became a college professor which pays the same regardless of which department you teach in.

#### HallsofIvy

##### Elite Member
Unfortunately, later, when I had a wife and kids it occurred to me that money should have come into it!

#### Subhotosh Khan

##### Super Moderator
Staff member
You know better than that. I am sure that you studied math for the beauty of it. Money nor application never came to mind!
Besides, Khan became a college professor which pays the same regardless of which department you teach in.
Jomo,

Either you are naïve or pretending to be naïve. You should know for most of the applied science professors, the majority of their income is derived from the "sponsorship" of industries (research grants). That grant is generally above and beyond the academic salary. I have been on both sides of that racket.

#### nasi112

##### Junior Member
A lot of people are here. Therefore, I will go and search for an empty relaxing thread

#### HallsofIvy

##### Elite Member
You could accomplish that by simply turning your computer off!