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At $8am$ particle $A$ is at point $(0,0)$ and moves horizontally to the right with constant velocity of $60km/h$. At the same time particle $B$ is at the point $(0, A+B+C+5)$ and moves horizontally to the left with a velocity of $30km/h$. what are the rates of change of the distance between the two particles at $8.00am, 9.00am$? and in general time $t$?

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What have you tried? –  littleO Nov 11 '12 at 10:59
    
$A$ and $B$ are names of particles, but then you add them as if they are numbers. What does that mean? And what's $C$? –  littleO Nov 11 '12 at 11:08
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1 Answer

let distance between cars be z

$z^2=x^2+y^2$

$z'=\frac{1}{z}({xx'+yy'})$

at time t the cars are moved a distance of

$30t+60t$

and

$y=A+B+C+5,y'=0,x=90t,x'=90$

$z'(t)=\frac{1}{z}(90(t)+0)$ where

$z=\sqrt{(A+B+C+5)^2+(90t)^2}$

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