Question

There are three parallel tracks of 120 km,between x and y.Train A starts at 9:00 am on the central track at speed of 60 km/hr from X.Train B starts at 10:00 am at speed of 60 km/hr from y.Train C starts at 11:00 am at 9 km/hr from x.------(continued)?

---------trains A and C are 100m each and train b is 150m long.
Q-at what time at what distance from x did the trains A and B meet each other?

Answer 1

Trains A and B meet 1.5 hours after A left station X. The meeting was 150 km from X.

Explanation:

We can ignore train C. It is not involved in the question. We also can ignore the length of the trains. You would consider that the trains have met when the engines have met.

The sum of the distances the 2 trains travel before meeting will be 120 km. Let's agree that time t starts when train A starts. The distances each train will have traveled at the time of the meeting:

Train A travels a distance given by `60 (km)/(hr)*t`

Train B is at rest for the first 1 hour of time t, therefore
train B travels a distance given by `60 (km)/(hr)*(t-1 hr)`

The sum of those 2 distance is 120 km. Therefore

`60 (km)/(hr) * t + 60 (km)/(hr)*(t-1 hr) = 120 km`

Expanding

`60 (km)/(hr)*t + 60 (km)/(hr)*t -60 (km)/cancel(hr)*1 cancel(hr) = 120 km`

Simplifying

`120 (km)/(hr)*t - 60 km = 120 km`

Solving for t

`120 (km)/(hr)*t = 120 km + 60 km`

`t = (180 cancel(km))/ (120 cancel(km)/(hr)) = 1.5 hrs`

The distance A has traveled is `60 (km)/cancel(hr)*1.5 cancel(hrs) = 150 km`

I hope this helps,
Steve

Answer 2

another approach?

Explanation:

Total distance is `120km`

For A;

It will cover `½` the distance before B would start the journey

since A have already started the journey , after one hour , A would be `60km` from `y`

Hence the distance left to cover is `60km`

at the same speed

A would cover `30km` so does B

Hence;

A covered `90km` from x

`T = 90/60 = 1.5hours` to reach the `90km` mark [ irrespective of their lengths ]