Question

bay city and vernonville are 184 mi. apart. a car leaves bay city traveling towards vernonville, and another car leaves vernonville at the same time,traveling bay city. the car leaving bay city avg. 10 mph more than the other, and they meet after 1 hr. and 36 min. what are the avg speeds of the cars

Answer 1

first establish the variables v = speed of car leaving bay city and v-10 is speed of ther other car
baycity car travels x miles and other car travels 184- x miles
time taken for each car is 1.6 hours

so v = x / 1.6
and v-10 = (184 - x) / 1.6
solve these system of equations and you have your answer

Answer 2

sorry i am doin it on scratch paper right now just a lil slow

Answer 3

is there another way of writing out the equation?

Answer 4

thats ok - i'm getting slow too - age i'm afraid - lol

Answer 5

x = 1.6v
v-10 = (184 - 1.6v) / 1.6
1.6v - 16 = 184 - 1.6v
3.2v = 200
v = 200/3.2 = 62.5 mph - car leaving bay city
other car travels at 52.5 mph

Answer 6

thanks you I appreciate it i got hung up on the equation but you cleared it up thanks!