The answer in this question is 44.7533 mphGiven information which we denote I as the distance of the automobile between the farmhouse, and S = the distance past the intersection of the highway and the road.
Then I^2 = 5^2 + s^2. Taking the derivative of both sides of this equation yields 2I(ds/dt) = 2s(ds/dt), so (dl/dt) = s(ds)/I(dt). When the automobile is 7 miles past the intersection we have;dl/dt = 55√(7/5^2 +7^2) and gives us the answer of approximately 44.7533 mph