Messages posted by uday_zingtudor

Hey, take x= -1

now can you do that???

The domain for "x" is R⁺

But then the diff. rules don' apply

~Cheerio!!!

Hi guyz,

If half the distance is taken as "x",

we have, x/60 + x/40 = 2x/k and we get k = 48. Now leave this!!!

We have avg.velocity. That means, the ratio of total distance travelled and the total time taken is 48.That's it!! So, we aint bothered about the total distance travelled and the total time taken. There's only one condition!!! Half the distance is to be travelled with a speed of 60 and the other half with 40!!! You aren't bothered with the distance travelled. You may travel an inch or a furlong!!

So, the book is wrong!!!

Stick for reality!!! Why do ya bother when there are easier, unquestionable methods???

~Cheerio!!!

You people are wrong about that distance and time!!!

Even if you travel for a distance of 480km and for 10hrs, the answer wud be the same.

Look, the question is only about the average velocity!!!

~Cheerio!!!

goiit_user_videos

Selling peacock fans by speaking many languages

Congratulations!!!

I think i'm the second person to wish you........after Sowmya ......(though not on goiit)

And............... this is my 500^th post!!!

~Cheerio!!!!!!

Hey ananth_pathri,

You can never ever burn Helium gas!!!

Definitely there would be relative curved motion 'coz the times of flight may be different. After one body comes to rest, the other continues its motion which is definitely curved!!

And also if they are projected one along z-axis and the other along x-axis, one wud absolutely find the relative motion to be curved.

If the case is only 2D, then while they are in motion,

horizontal motion:Both of them are moving with constant velocities and so the relative motion is linear

vertical motion : Both of them are movin' under the same accelaration .So in this direction, the relative motion would be linear

Now, when both the horizontal and vertical motion appear to be linear, the resultant motion wud also appear linear!!

~Cheerio!!!

This problem would be very helpful to you!!

http://www.goiit.com/posts/preList/51600/274971.htm#274971

67x²/2

Substitute pi/2+@ instead of @

Understood the mistake??

~cheerio!!!

Another way

$sinx=x-x{}^{3}/3!+x{}^{5}/5!-.......$

$tanx=x+x{}^{3}/3+2x{}^{5}/15+............$

Using these expansions u get the co-eff of x³term as 1/3 + 1/6 = 1/2

The other terms become equal to zero after dividing by x³ and then substuting the limit as zero.

So the ans is 1/2

~Cheerio!!!

http://www.school-for-champions.com/science/friction_coefficient.htm

http://jeffareid.net/real/ttstick.wmv

You are right man!!

But anchit is a good boy!! You've guessed wrongly abt him!!

And if I were permitted , i'd reveal some wierdest facts!!!

Why don't the institutes give the same key for jee and aieee when all of them give the so called "iit-jee coaching" and "aieee" coaching

The pointed one!!

Given that f(1) = 0

That implies the other root is c/a

~Cheerio!!

1=0

The max and min values of the equation acosx + bsinx + c are $c+\sqrt{a^{2}+b^2}$ and $c-\sqrt{a^{2}+b^2}$

Using this u get the answer 'C'

~Cheerio!!!

Write cot3x as 1/tan3x and cotx as 1/tanx

Now the expression reduces to the form 1+tan3xtanx/tan3x-tanx

And that 's equal to 1/tan2x = cot2x

~Cheerio!!!

Is the answer 786???

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