integration question (1 Viewer)

[J-Wang]

How do integrate this?

 

[nightweaver066]

 

[SpiralFlex]

Sorry on iPhone

-1/k int (-kv/(g-kv)) dv

Sorry on iPhone

-1/k int (-kv/(g-kv)) dv

-1/k int (g-g-kv/(g-kv)) dv

Split then integrate

 

[Shadowdude]

How do integrate this?

with respect to what?

 

[karnbmx]

with respect to what?

LOL. I honestly think he wouldn't have a problem if it wasn't with respect to v.

 

[Shadowdude]

LOL. I honestly think he wouldn't have a problem if it wasn't with respect to v.

you never know though

 

[J-Wang]

with respect to v

you never know though

 

[Carrotsticks]

with respect to what?

I think it's very clear that he means with respect to V, as this is a very common integral in Resisted Motion problems where the resistance is mkv.

 

[Mr Slick]

I got -1/k [ v + g/k ln (g-kv) + c

is that the answer?

 

[Mr Slick]

You didn't integrate it

 

[J-Wang]

I got -1/k [ v + g/k ln (g-kv) + c

is that the answer?

I just did the integration as well and got that too. But I'm pretty sure the answers are wrong. You see, I did it where I expressed acceleration as vdv/dx before integrating. But the answers have for some reason just expressed acceleration as dv/dx. Hence, they must be wrong and so i'll say we r correct.

If anyone is wondering, the question is from the kings 2009 paper

 

[Mr Slick]

How do integrate this?

Are you in year 12?

 

[Mr Slick]

I just did the integration as well and got that too. But I'm pretty sure the answers are wrong. You see, I did it where I expressed acceleration as vdv/dx before integrating. But the answers have for some reason just expressed acceleration as dv/dx. Hence, they must be wrong and so i'll say we r correct.

If anyone is wondering, the question is from the kings 2009 paper

Nah that is the answer!

Post up the question ....

 

[Mr Slick]

bro it is the integration of 1/ g- kv duhhh (i looked at the question)


how did u get V on the top?

you use dv/dt

there is no v.dv/dt

 

[D94]

I just did the integration as well and got that too. But I'm pretty sure the answers are wrong. You see, I did it where I expressed acceleration as vdv/dx before integrating. But the answers have for some reason just expressed acceleration as dv/dx. Hence, they must be wrong and so i'll say we r correct.

If anyone is wondering, the question is from the kings 2009 paper

You don't need to use vdv/dx, it requires a time (t) in the expression, so you need dv/dt. The solutions use dv/dt, I think you just read it wrong. It's the integral of 1/(g-kv) wrt v

 

[J-Wang]

You don't need to use vdv/dx, it requires a time (t) in the expression, so you need dv/dt. The solutions use dv/dt, I think you just read it wrong. It's the integral of 1/(g-kv) wrt v

shit. that's awkward. that makes the question incredibly easy then

 

[nightweaver066]

You didn't integrate it

I doubt the OP had trouble actually integrating it but with getting it in to standard form.

If not, I apologise for the incorrect assumption.