Two cars start from the same point at the same time. One travels north at 25 mph, and the other travels east at 60 mph. How fast is the distance between them increasing at the end of 1 hr? Find D after 1 Hr, solve D^2=25^2+ 60^2

Answer 1

if the cars start at (0,0), then their positions at time t are

one: (0,25t)
two: (60t,0)

distance d:

d^2 = (60t)^2 + (25t)^2

when t=1,

d^2 = 3600+625 = 4225
d = 65

2d dd/dt = 2(60t)(60) + 2(25t)(25)
at t=1,

2(65) dd/dt = 7200+1250
dd/dt = 65 mph

Answer 2

simple way:

Let the time passed since they left be t hours
So distance traveled by northbound car is 25t miles
distance traveled by eastbound car is 60 t miles

Now make a sketch calling the distance between them D
D^2 = (25t)^2 + (60)t^2 = 4224t^2
D = √4224 t
dD/dt = √4224 = 65 mph

or (your way)

D^2 = x^2 + y^2
when t = 1 , x=60, y=25
D^2 = 3600 + 625 = 4224
D = √4224

2D dD/dt = 2x dx/dt + 2y dy/dt
dD/dt = (x dx/dt + y dy/dt)/D

when t=1 , D = √4224 , dx/dt = 60 , dy/dt = 25 , x=60 , y = 25

dD/dt = (60(60) + 25(25))/√4225
= 4225/√4225
= 65

Often the choice of variables will simplify or complicate your solution. Notice that my way not only gave us the answer more quickly, but it also pointed out an interesting fact.
The rate does not even depend on the time, the two cars would be separating at 65 mph at any given time of t.

Answer 3

Nice to see that we both do these in that simple way.

Answer 4

Nice catch on the constant speed of separation. I hadn't noticed that, and was vaguely surprised that the 65 popped up again at the end.

Answer 5

you had nothing about the angle of separation. can you please answer this question: two vehicles leave point c travelling at 60km/h the angle between the two points is 60 degrees how far apart will A and B be after 5 hours.