can someone help me with my algebra homework?

Question

a slow train leaves canberra at 9:17 am and arrives in goulburn at 12:02 pm. on the same day, the express leaves canberra at 9:56 am and arrives in goulburn at 11:36 am. at what time does the express pass the slow train if each is traveling at a constant speed?

i would appreciate it if someone could show me how to start this. thx!

Answer 1

slow train takes 3:45=225 min moving at 1/225 min ^-1
fast train takes 1:50=110 min moving at 1/110 min^-1
slow train has gone 39/225 of the distance when the fast train starts.
they cover 1-39/225=75/75-13/75=62/75 of the trip
to meet.
they are approaching at 1/110+1/225=45/4950+22/4950=67/4950
min ^-1
62/75/(23/4950)=(62/75)(4950/67)=61.07 min or i hr 1 min 4 sec

Answer 2

The slow train covers the distance to Goulburn in 2:45. The express takes 1:40 for the same distance. The slow train is traveling at a speed of Dist./2:45. (Change to minutes: Distance /165minutes.) The express travels at Distance/100 minutes. We need to find the time when D/165(t) = D/100(t+39).

Answer 3

caraberra ---------------------goulbourn
|-----------X dist--------------------|
Slow train takes 165 min = 12.02 - 9.17
Fast train takes 90 mins = 11.36 - 9.56

Slow train speed = X/165
Fast train speed = X/90

fast train starts after 39 mins so in that time slow train has covered 39*X/165 distance

So if consider after "t" time fast train overtakes slow train then the quation will be

39X/165 + tX/165 = tX/90
so
39/165 + t/165 = t/90
t(1/90 - 1/165) = 39/165
t*75/90 = 39
t = 39*90/75

t= 46.8 Mins
here is you answer

Well boss i had solved this puzzle for you... but i will suggest to do all these things by urself rather than solving them this way
you wont learn any thing
you should develop ur own skills

Answer 4

time is 21 minutes after starting of second train(faster one).

Answer 5

see answer below