A forest ranger is in a forest 2 miles from a straight road. A car is located c
ID: 3190299 • Letter: A
Question
A forest ranger is in a forest 2 miles from a straight road. A car is located c miles down the road. The forest ranger can walk 3 miles per hour in the forest and travel 5 miles per hour along the road. In the picture below, the forest ranger walks in a straight line to a point x unit from the end of the road on the left, and then along the road.
The total travel time for the ranger to get to the car is the sum of the travel time in the forest and the travel time on the road. Write the total time T as a function of x, using the letter c for the distance from the end of the road on the left to the car.
T(x)=______
Toward what point on the road should the ranger walk in order to minimize the travel time to the car if...
(a) c = 9 miles? (Numerical Answer ONLY)
x= ___
(b) c = 1/5 miles? Remember that the point must lie between the left end of the highway and the car. (Numerical Answer ONLY)
x=______
A forest ranger is in a forest 2 miles from a straight road. A car is located c miles down the road. The forest ranger can walk 3 miles per hour in the forest and travel 5 miles per hour along the road. In the picture below, the forest ranger walks in a straight line to a point x unit from the end of the road on the left, and then along the road. The total travel time for the ranger to get to the car is the sum of the travel time in the forest and the travel time on the road. Write the total time T as a function of x, using the letter c for the distance from the end of the road on the left to the car. T(x)=______ Toward what point on the road should the ranger walk in order to minimize the travel time to the car if... (a) c = 9 miles? (Numerical Answer ONLY) x= ___ (b) c = 1/5 miles? Remember that the point must lie between the left end of the highway and the car. (Numerical Answer ONLY) x=______Explanation / Answer
distance till road = sqrt(2^2 + x^2) = sqrt(4 + x^2) distance on road = c-x so total time taken = [sqrt(4 + x^2) / 3] + [(c-x)/5] for time minimum differenciate above equation and equate it to 0 a) for c = 9 by differenciating [sqrt(4 + x^2) / 3] + [(9-x)/5] we get x/6(sqrt(4 + x^2)) - 1/5 equating it to 0 we get x = 2/3 miles ( answer) b) c = 1/5 differenciating [sqrt(4 + x^2) / 3] + [((1/5)-x)/5] and equating it to 0 we get x = 2/3 but x should be less than 1/5 so he shud directly go to car so x = 1/5 miles (answer)
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