Two cars leave town 150 kilometres apart at the same time and travel toward each other. One car's rate is 18 kilometres per hour more than the others. If they meet in 3 hours, what is the rate of the faster car ?
V1 = X km/h, V2 = x + 18 km/h.
d1 + d2 = 150 km.
3x + 3(x+18) = 150, 6x + 54 = 150, 6x = 96, X = 16 km/h, x+18 = 16 + 18 = 34 km/h = V2 = Speed of the faster car.
To find the rate (or speed) of the faster car, we can set up an equation based on the information given.
Let's say the rate of the slower car is 'x' kilometers per hour. Since the rate of the faster car is 18 kilometers per hour more, its rate is 'x + 18' kilometers per hour.
The distance traveled by the slower car in 3 hours is:
Distance = Rate × Time = x km/h × 3 h = 3x km
And the distance traveled by the faster car in 3 hours is:
Distance = Rate × Time = (x + 18) km/h × 3 h = 3(x + 18) km
Since they are traveling towards each other, the total distance covered by both cars should add up to 150 kilometers:
Distance of slower car + Distance of faster car = 150 km
Substituting the distances we calculated earlier, the equation becomes:
3x + 3(x + 18) = 150
Now we can solve this equation to find the value of 'x' and hence the rate of the faster car.
3x + 3(x + 18) = 150
3x + 3x + 54 = 150
6x + 54 = 150
6x = 150 - 54
6x = 96
x = 96/6
x = 16
So, the rate of the slower car is 16 km/h. And since the rate of the faster car is 'x + 18', the rate of the faster car is:
16 + 18 = 34 km/h
Therefore, the rate of the faster car is 34 kilometers per hour.