Question

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Automobile traffic passes a point P on a road of width w feet (ft) at an average rate of R vehicles per second. Although the arrival of automobiles is irregular, traffic until there is a gap in traffic of at leastt seconds is approximately T te engineers have found that the average waiting time a speed of 3.5 ft/s requires t = w/3.5 s to cross the road. Therefore, the average time the pedestrian will have to wait before crossing is f(w, R) (w/3.5)ewR/3.5 s. (Round your answers to one decimal place.) seconds. A pedestrian walking at 30 ft and R= 0.22 vehicle per second? (a) What is the pedestrian's average waiting time if w 56.5 (b) Use the linear approximation to estimate the increase in waiting time if is increased to 31 ft. 5,4 S (c) Estimate the change in waiting time if the width is increased to 32 ft and R decreases to 0.20. 55.3 What is the new estimated waiting time? (d) What is the rate of increase in waiting time per 1-ft increase in width when w = 40 and R = 0.39 vehicle per second? S

Answer 1

thnt tee anerge time the pelestinn thaHic is aWe asl wil haue to wait to a t-m wR/3.5 flu,R) we 25 the values and R= O22, we w 30 30 d30x0.22)/3 S 3.5 H20, 0.22) saconds こ56.5 ine appoimaton The at (w,R)= (30, 0. 22) is tu(20, O.22) Au f0,022) () AR - we rartit dvivatines omput t ule and he Produt the Choin Yule me wR/3. wK/3S wR/3.S 1 +wR R t we 3.5 3.5 3-5 3.5 the Chain Jule we wR/3.5 plR/3.5 - 3 3.5 At the Parint w(30, 0.22)4 ( 20,0 22) we fame frl30, 0 22) 464-2 (2) thre desivatines Aw= 31-3 o = AR O and in we Det Af Af h S pevmos ts tituti tee oleivati'res in () with Aw- 32-30-2 we in the lineau AR O.20 O.22 and = -0.02 to ostain appoximaton ミ5.4 X 2+484.2x(-002) time is inereared by approx eonda aimetiged wig f30, 0.22) t 1 56.5+ 57.6 Aeio nds That is, the time is Using pat La),

7/ time per ease in width, whsw df (40, 02). Ve in part me have d) The hate ine mircare in foot and R= te deuivatve obtineA 39 O . w (40xO +40x0.39 3 5 t(40, 0.39)= eOxo 3)/2-5 3 S