" A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck traveling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other and the truck leaves the intersection after 2 seconds, how fast are the car and the truck separating when the truck leaves the intersection? "

Here is one way.

After some time t seconds:

>>>the car travelled 30*t ft, so it is now (120 -30t) ft from the intersection.

>>>the truck travelled 40*t ft, so it is now 40t ft from the intersection.

Let x = distance between the car and truck at any time, in ft.

Imagine the figure,

It is a right triangle, with these:

hypotenuse = x

one leg = 120 -30t

the other leg = 40t

So, by Pythagorean theorem,

x^2 = (120 -30t)^2 +(40t)^2 ----(i)

Differentiate both sides of (i) with respect to time t,

2x(dx/dt) = 2(120 -30t)(-30) +2(40t)(40)

2x(dx/dt) = -7200 +1800t +3200t

2x(dx/dt) = 5000t -7200 ---------------(ii)

"...and the truck leaves the intersection after 2 seconds, how fast are the car and the truck separating when the truck leaves the intersection?"

That means, when t=2sec, what is dx/dt?

In (ii), we know t=2, but we don't now x.

In (i) we can get x when t=2, so,

x^2 = (120 -30t)^2 +(40t)^2 ----(i)

x^2 = (120 -30*2)^2 +(40*2)^2

x^2 = 60^2 +80^2

[if you know the 3-4-5 right triangle, you can say here now that x=100.]

x^2 = 3600 +6400 = 10000

x = sqrt(10000) = 100 ft

Substitute those into (ii),

2x(dx/dt) = 5000t -7200 ---------------(ii)

2(100)(dx/dt) = 5000(2) -7200

200(dx/dt) = 2800

dx/dt = 2800/200 = 14 ft/sec ------***

Therefore, at that time, the car and truck are separating at the rate of 14 ft/sec. ---answer.