A car leaves an intersection traveling east. Its position t sec later is given by x = t2 + t ft

intersection-travel

#1

A car leaves an intersection traveling east. Its position t sec later is given by
x = t2 + t ft. At the same time, another car leaves the same intersection heading north, traveling y = t2 + 4t ft in t sec. Find the rate at which the distance between the two cars will be changing 5 sec later. (Round your answer to one decimal place.)

Answer:

just wondering, is this 2t or t^2. Using t^2 for t2 you can plug in t=5 for t in both x and y equations x(5) = (5)^2+5=30 y(5) = (5)^2 + 4(5) = 45. use Pythagorean theorem to get a^2+b^2=c^2 ___ sqrt(45^2 + 30^2) = c. ___ c = 54.083 now rate with respect to time would be v = distance/time ___ 54.083 ft /5 sec = 10.82 ft/sec.