Application of Calculus (1 Viewer)

[fullonoob]

A 4-wheel drive vehicle which can average 40km/h and 110km/h off and on the road respectively, is 30km from the nearest point N on a straight highway. Which direction should the vehicle take to reach town T, 100km along the highway from N, in the least time? How long does the journey take?
By direction it means angle. I've forgot how to do this, but i remember the car will travel off the road then on the road for the least time. :apig:

 

[fullonoob]

Dw question answered. If you wna know the solution put a post here

 

[hello-there]

Dw question answered. If you wna know the solution put a post here

yer post it up noobie

also draw it up for me

 

[fullonoob]

nu uh i dont wna draw >_>
You'd have to figure how to draw it yourself, jsut relate it to my answer
Since TN = 100 make a point between T and N and call that P or whatever you want, PN will equal to x

t = d/ v
t = f(x) , where x is ditsance
t= t1 + t2
t= CP/ 40 + (100-x)/110
CP = root(900+ x^2)
t = root(900+x^2)/40 + (100-x)/110
t' = x(900+x^2)^-.5 / 40 - 1/110
t" > 0 therefore min tp
t' = 0
0 = x(900+x^2)^-.5 / 40 - 1/110
x^2 = 14400/105, skipped algebra cos cbf typing up
= 11.71 , positive since d cannot be <0
now you have your x distance
tan@ = 11.7/30
@ = 21' 19'

For the time you just use t = d/v since you already know distances , just a matter of subbing in

+1 rep