Question

Thorsten the geologist is in the desert, 10 km from a long, straight road. On the road, Thorsten's jeep can do 50kph, but in the desert sands, it can manage only 30kph. How many minutes will it take Thorsten to drive through the desert? (See details).

Thorsten is very thirsty, and knows that there is a gas station 25 km down the road (from the nearest point N on the road) that has ice-cold Cola.

A diagram of the problem:

(a) How many minutes will it take for Thorsten to drive to P through the desert?
Rate = Distance/time
Time = Distance/rate
Time = PA/30 kph
PA can be solved with Pythagorean theorem.

(b) Would it be faster if Thorsten first drove to N and then used the road P?
Found the time to get to N from A, then the time from N to P using the same techniques above.

(c) Find an even faster route for Thorsten to follow. Is your route the fastest possible?
???

Answer 1

(a) `54` minutes; (b) `50` minutes and (c) `3.7` km. from N it would take `46.89` minutes.

Explanation:

(a) As `NA=10km.` and `NP` is `25km.`

`PA=sqrt(10^2+25^2)=sqrt(100+625)=sqrt725=26.926km.`

and it will take `26.962/30=0.89873hrs.`

or `0.89873xx60=53.924min.` say `54` minutes.

(b) If Thorsten first drove to N and then used the road P,

he will take `10/30+25/50=1/3+1/2=5/6` hours or `50` minutes

and he will be faster.

(c) Let us assume he directly reaches `x` km. from `N` at S,

then `AS=sqrt(100+x^2)` and `SP=25-x` and time taken is

`sqrt(100+x^2)/30+(25-x)/50`

To find extrema, let us differentiate w.r.t. `x` and put it equal to zero. We get

`1/30xx1/(2sqrt(100+x^2))xx2x-1/50=0`

or `x/(30sqrt(100+x^2))=1/50`

or `sqrt(100+x^2)=(5x)/3` and squaring

`100+x^2=25/3x^2`

i.e. `22/3x^2=100` or `x^2=300/22` and

`x=sqrt(300/22)=3.7` km.

and time taken will be `sqrt(100+3.7^2)/30+(25-3.7)/50`

= `0.78142hrs.=46.89` minutes.