lewiston and vervnonville are 208 miles apart. a car leaves lewiston traveling towards vernonvilld, and another car leaves vernonville at the same time, traveling towards lewiston. the car leavving lewiston averages 10 miles per hous more than the other, and they meet after 1 hour and 36 mins. what are the average speeds of the cars?
+130511 Mmmmmm....that's a pretty good problem.....let's let 1 hr and 36 minutes be = 1.6 hrs.
Note that the total distance traveled by the cars combined = 208 miles ... and they travel for the same amount of time
And rate x time = distance
Let x be the rate of the slower car....so x + 10 is the rate of the faster car ... so we have
x(1.6) + (x+10)(1.6) = 208 simplify
1.6x + 1.6x + 16 = 208
3.2x + 16 = 208 subtract 16 from both sides
3.2x = 192 divide both sides by 3.2
x = 60 mph and that's the slower car's rate
And the other car travels at 10 mph more = 70 mph
+130511 Mmmmmm....that's a pretty good problem.....let's let 1 hr and 36 minutes be = 1.6 hrs.
Note that the total distance traveled by the cars combined = 208 miles ... and they travel for the same amount of time
And rate x time = distance
Let x be the rate of the slower car....so x + 10 is the rate of the faster car ... so we have
x(1.6) + (x+10)(1.6) = 208 simplify
1.6x + 1.6x + 16 = 208
3.2x + 16 = 208 subtract 16 from both sides
3.2x = 192 divide both sides by 3.2
x = 60 mph and that's the slower car's rate
And the other car travels at 10 mph more = 70 mph
