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Please can someone help me !

A fast train leaves Manchester for London, a journey of 330km, at 12 noon. A slow train, travelling half as fast, leaves London for Manchester at the same time. They pass each other at 2 p.m. Find the speed of each train.

Can someone please tell me the 2 equations needed for this question

Thanks a lot! :smile:
You could start by reasoning that, when the trains meet, the distances travelled by each of them will be in the ratio 2:1. The total distance is 330km and the elapsed time until the trains meet is two hours (both given).
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Avatar for MastaV
MastaV
OP
7
Thanks for the help, I was just stuck at trying to formulae 2 equations for it since this activity is about simultaneous equations
Reply 3
Avatar for luq_ali
luq_ali
21
Okay-this is very interesting-when I first saw this problem, I decided to quickly run and ignore it! :smile: However, I found it interesting...so perhaps while I don't know the answer, we can work through it together.

330KM
A Train (M)____________330K______________________________B (L)Train
(M is Manchester train and L is the London train, so A leaves Manchester and B is leaving London)

2hrs is the meeting time of the two trains...what I have so far is:

2X(the double speed train)X2(representing the 2hrs of travel)=1x X2 (with one x being the slow train at half speed, and again the 2 being the 2hours at that rate of speed.

Hope this is helpful.:smile:

Original post by MastaV
A fast train leaves Manchester for London, a journey of 330km, at 12 noon. A slow train, travelling half as fast, leaves London for Manchester at the same time. They pass each other at 2 p.m. Find the speed of each train.

Can someone please tell me the 2 equations needed for this question

Thanks a lot! :smile:
Original post by MastaV
Thanks for the help, I was just stuck at trying to formulae 2 equations for it since this activity is about simultaneous equations


You could set up equations for the distances of the two trains from Manchester in terms of velocity and time. These would be s = vt for the fast train and s= 330 - (1/2)vt for the slow train. Note the initial offset of 330km and the negative velocity for the slow train. They meet when they are instantaneously the same distance from Manchester.

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