Calculus Question (Related Rates)

Alright, so first off, i just wanna say that this isn't homework, im studying for a test, so no sass please! :)


anyways, the questions reads as follows.

an automobile travelling at a rate of 30 ft/s is approaching an intersection. When the automobile is 120 ft from the intersection, a truck travelling at the rate of 40 ft/s crosses the intersection. The automobile and the truck are on roads that are at right angles to each other. How fast are the automobile and the truck separating 2 seconds after the truck leaves the intersection?

so i know that i have to do the Pythagorean theory ... this is what i have so far


At t = 2, the auto is 60 ft from the intersection and the truck is 80 ft from the intersection.

60^2 + 80^2 = x^2
3600 + 6400 = x^2
10000 = x^2
x = 100 ft

a^2 + b^2 = c^2
2a da/dt + 2b db/dt = 2c dc/dt

then, i got stuck, so i looked at the solution...

2(60)(-30) + 2(80)(40) = 2(100) dc/dt
-3600 + 6400 = 200 dc/dt
2800 = 200 dc/dt
14 = dc/dt

The truck and the auto are separating at 14 ft/s.


what i dont undeerstand, is where exactly does the -30 come from, as well as where the 40 comes from. so if someone could just explain for me, that would be great!

1.) Automobile's Distance, A = 120 ft.
Automobile's Speed, dA/dt = - 30 ft./sec. (Negative because the distance is decreasing)
Truck's Distance = T
Truck's Speed, dT/dt = 40 ft./sec.
Time, t = 2 secs.
Distance Between = d

Find dd/dt:

Since the automobile has traveled some distance in 2 secs.,

A = A - [(dA/dt)]t (Absolute value because the direction here is irrelevant)
A = 120 - (30)(2)
A = 120 - 60
A = 60

T = (dT/dt)t
T = (40)(2)
T = 80

d² = A² + T²
d² = (60)² + (80)²
d² = 3600 + 6400
d² = 10000
d = √10000
d = 100 (Negative root disregarded)

Differentiate Implilcitly Over Time: