two cars are approaching an intersection.one is 5 miles north of intersection & is moving at a constant speed?

Question

of 40 miles per hour.at the same time other car is 6 miles west of the intersection & is moving at a constant speed of 30 miles per hour. Express the distance d between the cars as function of time t .

answer is SR( 2500t^2 - 360t + 13 )
pls. explain SR : square root

Answer 1

Let x=east, y=north, time=t in hours
The distance of the first car from the intersection as a function of time
= 5-40t
The distance of the second car from the intersectionas a function of time t
= 6+30t

As the two cars are travelling at 90 degrees of each other, they form a right triangle with the intersection, therefore using Pythagoras theorem,
D=sqrt(x^2+y^2)
=sqrt((5-40t)^2+(-6+30t)^2)
=sqrt(2500t^2-760t+61)

Note that the answer corresponds to the question you asked.

Your given answer corresponds to distances from the intersection of 3 miles (instead of 5) and 2 miles (instead of 6) from the intersection.

Answer 2

Pythagorean formula is used. Are you sure that the equation is correct? At t = 0 aren't the cars located on a triangle with sides 5 and 6? If so, then the distance between them is sqrt(5^2+6^2) which is not 13. I must not understand the geometry.

Answer 3

t=0 automobiles are 4 mile south and 5 miles east t+a million automobiles are how a approaches south and east ( bit much less, 60mph for a million sec and 50mph for a million sec) t+2 automobiles are how a approaches.....and so on t+x automobiles injury into one yet another or omit and are north and west of the junction