Question

A train leaves a train station at 1:00 PM .It travels at an average rate of 60mi/h.A high speed train leaves the same station an hour later.It travels at an average rate of 96mi/h.The second train follows the same route as the first train on a track parallel to the first. In how many hours will the second train catch up with the first train

Answer 1

OK, so our fundamental equation is going to be speed = distance/time
ie, distance = speed * time

When the trains catch up with each other, their "distance"s are equal, so we need to come up with an expression for speed * time, for both trains A and B.

Say, then, that the time after 1pm which the trains catch each other is t.

Therefore, your speed * time value for the first train is 60*t.

Since the second train leaves an hour later, it's time can be seen as t-1 hours after 1pm (when the trains meet).

So the second train's function is 96(t-1)

Now all you need to do is equate

60t = 96(t-1)
60t = 96t - 96
96 = 36t
t = 2 and 2/3 hrs = 2hrs 40min (after 1pm)

Since this time is after 1pm, and the second train left an hour later, it will take 1hr and 40min for the second train to catch the first.

I think that's right, but let me know if you have a different answer :)

Answer 2

Erm OK not sure why that squashed up everything I wrote... let me know if the equations etc are unclear :/

Answer 3

should read
\\ 60t = 96(t-1)
\\ 60t = 96t - 96
\\ 96 = 36t
\\ t = 2 and 2/3 hrs = 2hrs 40min (after 1pm)