Two cars start from the same place at the same time. After two hours, one car has traveled 30% farther than the other. If one car's speed is 20mph larger than the other car's speed, what are the speeds of the cars? Solve algebraically.
V2 = 1.3 V1
V2 = V1 + 20
Solve simultaneously.
The first step might be
0 = 0.3 V1 -20
Solve for V1; then V2
how did you get 1.3?
To solve this algebraically, let's assume the speed of one car is x mph. According to the problem, the other car's speed is 20 mph higher, which means its speed is (x + 20) mph.
We know that distance equals speed multiplied by time. After two hours, the distance traveled by the first car is 2x miles, and the distance traveled by the second car is 2(x + 20) miles.
Now, the problem states that one car has traveled 30% farther than the other car after two hours. Mathematically, we can express this as:
2(x + 20) = 2x + 0.3 * 2x
Simplifying this equation, we can remove the parentheses:
2x + 40 = 2x + 0.6x
Combining like terms, we have:
2x - 2x + 0.6x = 40
0.6x = 40
Now, we can solve for x by dividing both sides of the equation by 0.6:
x = 40 / 0.6
x ≈ 66.67
Therefore, one car's speed is approximately 66.67 mph. Since the other car's speed is 20 mph higher, its speed is:
66.67 + 20 = 86.67 mph
Hence, the speeds of the cars are approximately 66.67 mph and 86.67 mph.