How will i solve this problem.A train leaves point A at 5 am and reaches point B at 9 am. Another train leaves point B at 7 am and reaches point A at 10:30 am.When will the two trains meet ? Ans 56 min
Let speed of A=x and that of B = y (mph) B does journey in 7/8 of the time that A does. Hence y=(8/7)x
A takes 4hr so AB=4x Let t= time A has travelled when they meet. So B has travelled t-2 hr
When they meet distance gone by A + distance gone by B = AB=4x
Hence xt+y(t-2)=4x But y=(8/7)x Sub for y and solve for t
Problems like this can be solved graphically.
Put down a pair of co-ordinate axes with time along the x-axis and distance up the y-axis, and let A have a (y) co-ordinate of 0 and B a (y) co-ordinate of 1.
The motion of the first train is then described by a straight line running from (5,0) to (9,1).
The motion of the second train is described by a straight line from (7,1) to (10.5,0).
The trains meet at the point where the lines cross.
The intersection point can be calculated analytically.
The equation of the first line will be
$\displaystyle y=\frac{x-5}{4}$
and the second line
$\displaystyle y=-\frac{1}{3.5}(x-10.5).$
Equate the two values for $\displaystyle y$ and solve.
"train A" started at 5 AM and "train B" started at 7 Am to B started 2 hours after A. However, I don't like biffboy's notation. The problem says that A is a starting point, not one of the trains! And it is not a good idea to use a letter to represent two different things. His method, of course, is correct.
Here is yet another way to do this problem. Let x the distance between A and B, in km., rather than the speed of one train. Since the first train take 9- 5= 4 hours to make the trip, it has speed x/4 km/hr. The second train take 10:30- 7= 3.5 hours to go the same distance so its speed is x/3.5 km/hr.
Let t be the number of hours after 5 AM when they meet. The first train has traveled for t hours and so will have gone (x/4)t km. It will be (x/4)t km from point A. The second train, leaving B at 7 AM, two hours after 5 AM, will have been traveling for, as before, t- 2 hours and will have gone (x/3.5)(t- 2) km. it will be (x/3.5)t km from point B. Since the two trains have met, the total distance they have gone must be x km: (x/4)t+ (x/3.5)(t- 2)= x. The first thing we can do is divide through by x: t/4+ (t-2)/3.5= 1. Solve that for t and use t (the number of hours after 5 AM that they met) to find the time they met.
assign a rate of 50 mph for train leaving A. Distance Ato B -200miles
Train from B 200/3.5=57.1 mph
@ 7Am train is 100 miles from B
Trains approach at 107.1 mph Time =100/107.1 =.934 hr = 56 min
Okay, that's clever! Since neither distance nor speeds were given, it is reasonable to believe that they are irrelevant and choose a specific value for one of those (the others, the other speed and distance as bjhopper did, or the two speeds if you choose specific distance).
I do have three comments. 1) The problem asks "When will the two trains meet". The answer should be a clock time not "56 minutes". More important,
2) bjhopper does not seem to have taken into account the fact that the two trains left at different times.
3) The fact that the distance from A to B was 200 miles means that, at a "closing speed" of 107.1, they must close the entire 200 mi between then- it should be 200/107.1 not 100/107.1.
At 7AM the train which left A at 5AM is 100 miles from B.The train at B is leaving.The time of day that theymeet is 7.56 AM. I did not show this because the OP gave an answer in minutes. Iam a horrible typist and frequently lose messages before posting.Sone very long.I like to post ASAP