Question

A train 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?

Answer 1

Answer:

9:00 P.M.

Step-by-step explanation:

#1. Let x= the time in which the second train will catch up with the first

#2. Since x is the time in which the second train will catch up with the first and 3 P.M. is the time in which the second train left, we subtract that from x multiplied by the rate the second train travels, and in this case, is 60 mph= the rate in which the first train is traveling.

: 60(x-3)= 40x

: 60x-180= 40x

Subtract 60x on both sides to equal

: -180= -20x

Divide -20 on both sides to equal

: 9=x, which is 9 P.M.

Glad it helped! :)

Answer 2

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

$40t =60(t-3)$

distributing 60 over t-3

$40t =60t-180$

$60t -40t=180$

$20t =180$

$t =180/20$

t=9

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