Question

Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 10 miles per hour. At the same time, the other car is 5 miles east of the intersection and is moving at a constant speed of 20 miles per hour. (a) Express the distance d between the cars as a function of time t. (Hint: At t= 0, the cars are 2 miles south and 5 miles east of the intersection, respectively.) d = 500+? - 240 t + 29 (b) Use a graphing utility to graph d =d(t). For what value of t is d smallest? t~ hours (Round to three decimal places.)

Answer 1

Sol' Speeds aue V = 10 mph V₂ = 20mph kimile E w I 2 mild 2 mill I ca The initial distance b/w the caus is = V2+52 as = 124 mile. But the cars are moving so the sides of our right trångle are changing for can 1, Distance b/w car and intersection point equals (2-lot) where tis in hours. for Can 2, Distance b/w can & inter section point equals (5-20t). So, the distance b/w caus D = (2-10t)2 + (5-20+)2 DE 14+100t2 40t +25 +4006-2000 To = √500t? 240 t +29 | This is your distance function

b Differentiate distance function 4 put it zero, to find value of + for smallest d. D = (soot-2404+29) *(100-240) so ! (500t - 240 t +29) (100-240) = 0 ort. 1oot - 240 =0 t = 240 100 It=2.4 hour not applicabt or (500t? - 240 t +2972 =0 On solving, t = 0.24 hour = 14.4 min) Ans.