a plane traveling at 400 mph leaves the philadelphia airport at 1pm heading due east. two hours later a second plane traveling at 600 mph leaves the airport heading in the same direction. at what time does the second plane overtake the first?
When the 2nd overtakes the first, they will have travelled the same distance:
let the time of the first plane since 1:00 pm be t hrs
distance covered by first plane = 400t miles
let the time of the 2nd plane be t-2 hrs
distance covered by 2nd plane = 600(t-2)
400t = 600(t-2)
400t = 600t - 1200
-200t = -1200
t = 6
so time since 1:00 is 6 hrs, or the time will be 7:00 pm
To find out at what time the second plane overtakes the first plane, we need to determine the time it takes for the second plane to catch up with the first plane.
Let's assume the time it takes for the second plane to catch up is "t" hours.
The first plane travels for a total of t + 2 hours (since it departed two hours earlier).
Since the first plane is traveling at a speed of 400 mph and the second plane is flying at 600 mph, the second plane covers 600t miles during this time, while the first plane covers 400(t + 2) miles.
To overtake the first plane, the second plane needs to cover the same distance. Therefore, we can set up the following equation:
600t = 400(t + 2)
Now, let's solve this equation to find the value of "t":
600t = 400t + 800
600t - 400t = 800
200t = 800
t = 800/200
t = 4
So, it will take the second plane 4 hours to catch up with the first plane.
Since the second plane departs two hours after the first plane (at 1pm + 2 hours = 3pm), we can find the time when the second plane overtakes the first plane:
3pm + 4 hours = 7pm
Therefore, the second plane will overtake the first plane at 7pm.