Weird math problem (1 Viewer)

[iStudent]

This is a MX1 problem, but I came across this while doing Terry Lee's 4u book...

if , and initially m/s at , find v in terms of x and describe its resultant motion

Textbook's solution:








intiially, and

hence,



after simplifying and rearranging, you get this:

Compare this to what I did:







intially, and

hence,





after simplifying and rearranging, you get this:

as you can see, you get a plus sign which is different to the textbook method

Someone explain please

 

[QZP]

Ran into this problem a while back too. Would like to know reasoning!!

 

[Kurosaki]

I think it has something to do with the absolute value sign you've imposed.

 

[aDimitri]

It's something to do with the abs value that we impose when we integrate. If you leave it as:

Then you get the same solution as the textbook.

 

[iStudent]

But you can't log a negative :/

 

[Squar3root]

why did you decide to multiply both sides by -1 for no reason?

 

[iStudent]

I felt like LHS was easier to integrate this way. (just a habit)
it's still mathematically correct though

 

[Squar3root]

I felt like LHS was easier to integrate this way. (just a habit)
it's still mathematically correct though

I may be wrong; but when you multiply both sides by -1 you also change the sign of the constant when you're integrating

 

[iStudent]

I may be wrong; but when you multiply both sides by -1 you also change the sign of the constant when you're integrating

I don't think that matters.. :/ because it's a constant.
I think the absolute value sign when integrating needs to be reworked.

 

[Squar3root]

I don't think that matters.. :/ because it's a constant.
I think the absolute value sign when integrating needs to be reworked.

http://www.hsccoaching.com/documents/24.html

call him and ask whats up

 

[iStudent]

http://www.hsccoaching.com/documents/24.html

call him and ask whats up

That will be my last resort

 

[Squar3root]

That will be my last resort

maybe he made a mistake. can check here: http://www.hsccoaching.com/documents/35.html

 

[QZP]

maybe he made a mistake. can check here: http://www.hsccoaching.com/documents/35.html

Just look at the math lol... this has nothing to do with T.Lee or some textbook. OP found a contradiction in two different, (seemingly) valid methods and that's the problem

 

[Squar3root]

Just look at the math lol... this has nothing to do with T.Lee or some textbook. OP found a contradiction in two different, (seemingly) valid methods and that's the problem

OP did mention (in his first line) that it is out of TL's book lol :/ and i was thinking maybe TL made an error (unlikely)

 

[braintic]

When you do the rearrangment (the bit you didn't show), why didn't you get a +/- due to the absolute value?
You then simply have to choose the correct sign to match the initial conditions. That makes it a minus.

 

[aDimitri]

But you can't log a negative :/

You also can't square root -1, but we still have i don't we? :S

 

[Squar3root]

You also can't square root -1, but we still have i don't we? :S

huehuehue

 

[Kurosaki]

But you can't log a negative :/

Euler's identity says otherwise =)

 

[Squar3root]

But you can't log a negative :/

Euler's identity says otherwise =)

log(-1) = i*pi

 

[braintic]

log(-1) = i*pi

Just wondering, other than the fact that it is mathematically consistent, can anyone actually visualize this?
What does it really mean to say that e^(i.pi) = -1 ??
Does anyone have some insight which goes beyond just doing mindless algebra?