I have poor integration skills, give me a hand (1 Viewer)

[lacklustre]

Hiya.

Just lazing about doing some motion questions.

Need to integrate the following:

Int. 1 + log_ex dx

Cheerios.

 

[foram]

Hiya.

Just lazing about doing some motion questions.

Need to integrate the following:

Int. 1 + log_ex dx

Cheerios.

lol, you need to memorise this result i think. I've forgotten it.

actually, it's xlnx i think

yes int. 1+ lnx

= xlnx + C

 

[lacklustre]

lol, you need to memorise this result i think. I've forgotten it.

actually, it's xlnx i think

yes int. 1+ lnx

= xlnx + C

Yeah i did have a feeling it was supposed to be memorized. I looked at Cambrdige and apparently int. Inx = xInx - x + c. Thanks you're right foram.

Came across another question I couldn't do:

A plane lands on a runway at 100m/s. It then brakes with a constant deceleration until it stops 2km down the runway.

a) explain why the equation of motion is a = -k, for some postive constant k. (because it is decelerating right?)By integrating with respect to x, find k, and find v² as a function of x.

 

[tommykins]

lol, you need to memorise this result i think. I've forgotten it.

actually, it's xlnx i think

yes int. 1+ lnx

= xlnx + C

For proof -

let u = x, u' = 1, v = lnx, v' = 1/x

dy/dx = vu' + uv' = lnx + x*(1/x) = lnx + 1.

 

[vds700]

Yeah i did have a feeling it was supposed to be memorized. I looked at Cambrdige and apparently int. Inx = xInx - x + c. Thanks you're right foram.

Came across another question I couldn't do:

A plane lands on a runway at 100m/s. It then brakes with a constant deceleration until it stops 2km down the runway.

a) explain why the equation of motion is a = -k, for some postive constant k. (because it is decelerating right?)By integrating with respect to x, find k, and find v² as a function of x.

a = -k
d/dx (1/2 V^2) = -k
(1/2)v^2 = I -kdx
=-kx +c
when x = 0, v = 100
therefore c = 5000
(1/2)v^2 = -kx + 5000
when x = 2000, v = 0, therefore k = 5/2
(1/2)v^2 = (-5x/2) + 5000
v^2 = -5x + 10000

 

[lacklustre]

a = -k
d/dx (1/2 V^2) = -k
(1/2)v^2 = I -kdx
=-kx +c
when x = 0, v = 100
therefore c = 5000
(1/2)v^2 = -kx + 5000
when x = 2000, v = 0, therefore k = 5/2
(1/2)v^2 = (-5x/2) + 5000
v^2 = -5x + 10000

I don't get the first half. This is my thinking:

d/dx(1/2 v²) = -k
1/2 v² = -kx + C
v² = -2kx + C

Then x = 0, v = 100: and you get C = 10 000.

I'm confused. So does that mean i have to leave it at 1/2 V² = -kx + C.......and then sub?

Tell me where i'm wrong.

 

[vds700]

I don't get the first half. This is my thinking:

d/dx(1/2 v²) = -k
1/2 v² = -kx + C
v² = -2kx + C

Then x = 0, v = 100: and you get C = 10 000.

I'm confused. So does that mean i have to leave it at 1/2 V² = -kx + C.......and then sub?

Tell me where i'm wrong.

yes i get confused too if i multiply by 2, i mean, do u make the c into 2c or what? So i always leave it as (1/2)v^2 and then sub to avoid any confusion. I reckon my answer is right. v^2 = 10000 - 5x. When x = 0, v = 100 and when x = 2000m, v = 0.

 

[namburger]

yes i get confused too if i multiply by 2, i mean, do u make the c into 2c or what? So i always leave it as (1/2)v^2 and then sub to avoid any confusion. I reckon my answer is right. v^2 = 10000 - 5x. When x = 0, v = 100 and when x = 2000m, v = 0.

It doesnt really matter if it were C or 2C, because you will get the same equation in the end.

Case 1
1/2 v² = -kx + c
v² = -2kx + C
let x = 0, v = 100
C in this case = 10000

v² = -2kx + 10000
at x = 2000, v = 0
0 = -4000k + 10000
k = 5/2
Sub this into v² = -2kx + 10000
v²= -5x + 10000

Case 2
1/2 v² = -kx + c
v² = -2kx + 2c
let x = 0, v = 100
c = 5000
v² = -2kx + 2c
v² = -2kx + 10000
Following the same process in case 1 to find K and you will get the same answer.