Math car distance problem?
Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00am. each heading for Wildwood. One car's average speed in 10 mph more than the other's. The fastest car arrives at Wildwood at 11:00 am, half an hour before the other car. What is the average speed of each car? How far did each travel?
Comments
3x = 3.5(x-10)
0.5x = 35
x = 70 mph, second car = 60 mph
distance = 3(70) = 210 miles
The slowest car's average speed is x. The difference in the distance they traveled in 3 hours is equal to the distance the slow car travels in half an hour.
3(x+10) - 3x = .5x
3x + 30 - 3x = .5x
.5x = 30
x = 60
So one car has an average speed of 60mph, and the other is 70mph. To get the distance, do 3*70 or 3.5*60, which is 210 miles.
ok that's user-friendly, the 4 hundred(x+5), that's the recent changed version of the engine. Plus the 1st area that's 4 hundred cases x=400x 400x+400x+2,000 that's 800x+2,000km in line with litre of gas.
3hr(10mi/hr + x mi/hr)=y mi
3.5x=y
30+3x=3.5x
x=60mi/hr for the first car
70mi/hr for the second.
3*70=210 mi.