Two cars are traveling on the same highway. The first car is traveling at an average rate of 44mi/h. The second car leaves an hour late traveling at an average rate of 55mi/h. How long will the cars have traveled when the second car catches up with the first car?
Thank You
To find out how long it will take for the second car to catch up with the first car, we can use the concept of relative speed.
Let's assume that the distance traveled by both cars when they meet is "d" miles.
The first car travels at an average rate of 44 mi/h, so after "t" hours of travel, it will have covered a distance of 44t miles.
The second car leaves an hour late, which means it will have been traveling for "t - 1" hours. Since it travels at an average rate of 55 mi/h, its distance traveled will be 55(t - 1) miles.
When the second car catches up with the first car, the distances covered by both cars will be the same. So, we can set up the equation:
44t = 55(t - 1)
Let's solve for "t":
44t = 55t - 55
11t = 55
t = 5
Therefore, it will take 5 hours for the second car to catch up with the first car.
To find out how far they would have traveled, we can substitute the value of "t" back into either of the equations above. Let's use the equation for the first car:
Distance covered = 44t = 44 * 5 = 220 miles.
So, when the second car catches up with the first car, both cars will have traveled a distance of 220 miles.
key concept: both would have covered the same distance
let time of first car until catchup be t hrs
let time of 2nd car until they meet be t-1 hrs
55(t-1) = 44t
55t - 55 = 44t
11t = 55
t = 5
distance = 5(44) or 4(55), same thing,
= 220 miles