Car 1 & Car 2 leave the same location . Car 1 leaves the location 10 minutes before Car 2 and is traveling 65 mph. Car 2 is traveling at 75 mph. How long before Car 2 catches Car 1?
car 1 is 1/6 * 65 = 65/6 miles ahead when car 2 starts.
car 2 is going 10 mi/hr faster, so it will take
(65/6)/10 = 65/60 hr to catch up. In other words, 65 minutes.
To find out how long it takes for Car 2 to catch Car 1, we can set up an equation based on their speeds.
Let's say that it takes "t" hours for Car 2 to catch Car 1.
During this time, Car 1 has been traveling for "t + 10" minutes (since it left 10 minutes earlier), which is equal to (t + 10)/60 hours.
Using the formula Distance = Speed * Time, we can set up the equation:
Distance covered by Car 2 = Distance covered by Car 1
75t = 65(t + 10)/60
Now, let's solve this equation to find the value of t:
75t = 65(t + 10)/60
Multiply both sides by 60 to get rid of the denominator:
4500t = 65(t + 10)
Distribute on the right side:
4500t = 65t + 650
Move all the terms with t to one side and the constants to the other:
4500t - 65t = 650
Combine like terms:
4435t = 650
Divide both sides by 4435:
t = 650/4435
So, Car 2 catches Car 1 after (650/4435) hours.
To find the time in minutes, we multiply this value by 60 since there are 60 minutes in an hour:
(650/4435) * 60 ≈ 8.70 minutes
Therefore, Car 2 catches Car 1 approximately 8.70 minutes after Car 1 leaves.