Physics, asked by aparajitaTata495, 1 year ago

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.

Answers

Answered by kvnmurty
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.

Attachments:

kvnmurty: click on thanks button above please
Answered by shashankabanerjee773
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)

Attachments:
Similar questions