Question

A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking? (Round your answer to two decimal places.)

Answer 1

4.98 ft/s Let's determine the distance between the man and the woman for the moment that she's been walking 15 minutes. For this you can create a right triangle where one leg is 500 ft long (the east west difference between their locations) and the other leg is (distance man walked for 20 minutes + distance woman walked for 15 minutes). So Distance man walked = 20 min * 60 s/min * 2 ft/s = 2400 ft. Distance woman walked = 15 min * 60 s/min * 3 ft/s = 2700 ft. So the north south different in the man and woman's location is 2400+2700 = 5100 ft and will be increasing by 5 ft/sec. Creating a function of time (in seconds) for the distance the two people are apart is f(t) = sqrt(500^2 + (5100 + 5t)^2) where t = number of seconds from the 15 minutes the woman has been walking. For rate of change, you want the first derivative of the function. So let's calculate it. f(t) = sqrt(500^2 + (5100 + 5t)^2) f(t) = sqrt((5100 + 5t)^2 + 250000) f'(t) = d/dt[ sqrt((5100 + 5t)^2 + 250000) ] f'(t) = 0.5((5t + 5100)^2 + 250000)^(-0.5) * d/dt[ (5t + 5100)^2 + 250000 ] f'(t) = d/dt[ (5t + 5100)^2 ] / (2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 2(5t + 5100) * d/dt[ 5x + 5100 ]/(2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 5(5t + 5100/sqrt((5t + 5100)^2 + 250000) f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) Now calculate f'(t) for t = 0. So f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) f'(0) = (25*0 + 25500)/sqrt((5*0 + 5100)^2 + 250000) f'(0) = 25500/sqrt((5100)^2 + 250000) f'(0) = 25500/sqrt(26010000 + 250000) f'(0) = 25500/sqrt(26260000) f'(0) = 25500/5124.45119 f'(0) = 4.976142626 ft/sec So the man and woman are moving away from each other at the rate of 4.98 ft/s.