Distance Between Two CarsA car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 27 ft from the intersection. If the speeds of the cars at that instant of time are 7 ft/sec and 14 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.) ------------------ft/s

A North O West intersection let's Consider the distance is the first Car from the intersection is I and the distance of the second Can from the entersection is y Given that the first can is travelling west it's position after A second is 21 st from the intersection . x = 21 Also the position of the second Can after 4 second is 27 ff from the intersection ". y = 27 ft .Using pythagorean theorem, we Can find the distance between two cars . ( we say it z ) Pythagorean theorem state that in a right Triangle OAB , the square of hypotenuse is equal to the sum of the squares of the others two legs . If Of and OB are the sides and AB is the hypotenuse of the triangle then OAN POB = ABY x 7 4 = zz [where from the picture we see 2 = 1 x 7 4 2 that OA = x OB = 4, AB = Z V ( 21 ) 7 27 ) = ) 2 = 1 441+729 7) 2 = 1 1170 the distance between two cars is Z = VITO ft Now to find the rate at which the distancebetween the two cars is . changing We need to differentiate the equation . x ty " = 2 " with respect time ( t ) because de . is the rate at which the distance between the two cars is changing . Also Here given that the speed of the first can is 7 ft / see and the speed of the second canis 14 ft / see We know that speed = distance time For the first can speed = dx It For the second Can speed = dy de - 7 ft / see ly If = 14 #7 / see Now xi py? = z~ Differentiate with respect to time ( t ) , of ( x 7 yr ) = o ( 27 )( x 2 ) pd = 4 ( 2 1 ) -) 2 2 dre If + 2y dy It I = 27 d2 = n ( fix ) ) " - 1 x- our X at * dy df dx = ) . x dx It ty dy = ) dz 21 X 7 + 27 X 14 1 1170 197 + 378 = V1ITO 5 25 = 15 . 3 ( By sounding to dz one decimal place = 15.3 ft / see Hence , The rate at which distance between the two cars is changing is 15.3 ft /s