From point 'A' located on a highway, one has to get by a car as soon as possible to point 'B' located in the field at a distance ' l ' , from point 'D' . if the car moves n times slower in the field at what distance x from D one must turn off the highway.
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Answered by
125
see the diagram please.
The car travels with a speed v on the field and with a speed v * n on the main highway. The car starts from A and travels on the road upto C and then travels along CB on the field till B.
Let the distance AD = d. The perpendicular distance of point B on the field from the highway is DB = L. Let the distance from point D of C (turning point) be x.
time taken for traveling distance AC = t1 = (d - x) / (v n )
time duration for travelling distance CB = t2 = √(L²+x²) / v
Total time duration = T = t1 + t2 = √(L² + x²) / v + (d - x) / (v n)
We want to minimize T wrt x, so differentiate and dT /dx = 0
dT/dx = 2 x / [2 v √(L² + x²) ] - 1 / (v n)
equating it to 0, we get √(L²+x²) = n x
=> x = L / (n² - 1) This is the answer.
Now we can confirm that T is minimum at this value of x by finding the second derivative of T wrt x.
d²T / dx² = [L² - x²] / [v (L² +x²)^1.5] > 0
as L > x as n > 1
Hence we get that at the distance x = L /(n² -1) from the point D, the car must turn in to the field, in order to reach there in the least amount of time.
The car travels with a speed v on the field and with a speed v * n on the main highway. The car starts from A and travels on the road upto C and then travels along CB on the field till B.
Let the distance AD = d. The perpendicular distance of point B on the field from the highway is DB = L. Let the distance from point D of C (turning point) be x.
time taken for traveling distance AC = t1 = (d - x) / (v n )
time duration for travelling distance CB = t2 = √(L²+x²) / v
Total time duration = T = t1 + t2 = √(L² + x²) / v + (d - x) / (v n)
We want to minimize T wrt x, so differentiate and dT /dx = 0
dT/dx = 2 x / [2 v √(L² + x²) ] - 1 / (v n)
equating it to 0, we get √(L²+x²) = n x
=> x = L / (n² - 1) This is the answer.
Now we can confirm that T is minimum at this value of x by finding the second derivative of T wrt x.
d²T / dx² = [L² - x²] / [v (L² +x²)^1.5] > 0
as L > x as n > 1
Hence we get that at the distance x = L /(n² -1) from the point D, the car must turn in to the field, in order to reach there in the least amount of time.
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kvnmurty:
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Answered by
15
Answer:x=L/√n^2-1
Explanation: Please look at the graph .
Let AD be L (distance on highway)
DB be l ( distance in field)
And CD be x
Distance between AC = L-x
After some distance the car will turn off the highway at point c and move along CB , until it reaches point B. Now CB = v\n.
(for slower velocity v\n)
Time = distance\time
T={L-x/v +n√x^2+l^2/v}. (CB=√x^2 + l^2)
1/v(L-x+n(l^2+x^2)^1/2)
Now using differentiation-
dT/dx=1/v(0-1+n1/2( l^2+x^2)^1/2.2x)
= 1/v{-1+nx/√l^2+x^2}
For minimum time -
dT/dx=0
-1+nx/(√l^2+x^2)=0
nx=(√l^2+x^2)
(n^2x^2)=( l^2+x^2)
(n^2-1) x^2=l^2
X= l/√ (n^2-1)
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