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ah you posted it again

Yup, i was staring it for hours i didn't even know how to do it -_-

I tried working backwards but it didn't work. Or using the pythagorus theorem *sigh*

that looks soo scary.

LOL, I know its killing me =/

|dw:1317388650593:dw|

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Like this :
3999.23 / dt = 2000a + 2000b / dt ?

|dw:1317388825437:dw|
this is the graph of time against distance. Very badly drawn.

\[H^2 = (100 - 30t)^2 +(100-40t)^2\]

\(t\) is in hours

Don't you find H as well ? That is 141.42 ?

from the graph, the slope is -0.333...
The equation of the line is t = (-0.0333...)d + 3.333...

SimonS is trying to explain what Ishaan is saying and exactly right.
So Mimi, what is B(t)?

and for the record what Ian is saying is unfortunately wrong.

B(t) is the distance B from the intersection ?

Yes

The distance from the intersection of the driver driving at 30 km/hr

Yup, do i expand (100-30t)^2 ?

No.

Or not yet. So A(t) = 100 - 40t and hence B(t) = ... ?

B(t) = 100-30t

Right. Good.

Then the pythagoras theorem xD

so d(t)^2 = A(t)^2 + B(t)^2
= ....
Now expand.

H^2 = 2500t^2 - 14000t + 20000
What do i do next ? differentiate it ?

you want H(t)= sqrt(stuff)
then differentiate, set to zero and solve for t

ah, in which case, the graph was right.

Can I differentiate it keeping h^2 ?

Let f(t) = h^2(t). Differentiate this function.

f(t) ?

won't make any difference differentiate with h^2 so it is easy

But if this causes anyone confusion, differentiate the function h(t)

h^2 = 5000t-1400
h = squareroot ( 5000t-1400)
Is that right >

No. As you write above:
H^2 = 2500t^2 - 14000t + 20000

Don't I differentiate it >

Yes, but you made a few mistakes and your notation is wrong.
(h^2) ' = 5000t - 14000

Okay, the its h' = squareroot (5000t-1400)

then*

No, no, definitely not.
(h^2) ' does not equal (h')^2

you are dropping a zero on 14000

Just set (h^2) ' = 0 and solve for t

i.e., (h^2) ' = 5000t - 14000 = 0
means t = ....
Now for that t, h(t) = ....

Oh I got it 2.8 thank you (: