Atrain left a station a station at noon going 40mph. a second train left at 3 p. m. going 60mph. at what time will the second train catch up with the first?

Here speed of first train is slower than speed of second train. So if they catch up , the distance traveled will be same since they left the same station at different times.

But the time traveled will be different .

Let us say the two trains meet at time t ,

Distance = speed * time

Distance traveled by first train = 40 t

Since second train is leaving after three hours , so

Distance traveled by second train = 60(t-3)

As discussed earlier, distance traveled will be same, so

distributing 60 over t-3

t=9

So the second train will meet the first at 9 p.m.

the answer is 48m^2!

step-by-step explanation:

answer: 0

step-by-step explanation: the positive, like 5, + its opposite, like -5 would be 0

5+(-5)=0, or 5-5=0