Time of Trip

[Guest]

A,B, and C start at the same time from R to travel to W, 40 miles away.

A walks at 1 mph.
B " " 2 mph.

C rides at 8 mph.

C travels to W, then back until he meets A, picks up and transports a certain distance, again travels back, picks up B, which he transports a distance to allow all to reach W at the same time. Calculate time of trip.


When C reaches W, A will be at D.

E = point where C picks up A.

EG = distance C carries A.

RF = distance traveled by B until C leaves A at G.

H = location where C picks up B.

FH = distance B walks.

HG + GW = distance C carries B.


A walks from G to W.
B walks from F to H, and carried from H to W.

The time from F to W = x hrs., which is the time A walks from G to W.


Illustration attached

Answer given: 15.24 hrs.

How do I proceed?

[Guest]

Subject should read: Time of Trip.
Sorry for typo.

[Guest]

How much of last post is the question and how much is your working out. Is the illustration part of the question or to illustrate your understanding of the question.

Are the first few lines up to "Calculate time of trip. " the only bit that is the question.

I don't understand why A is left to walk to W from G if he was already picked up and could have arrived at W with B and C when C completed the trip after picking up B.

If I assume once A is picked up C moves on and picks up B and continues to W. There is no mention in the question that C can only cope with 1 passenger.

Normally ........
A @ 1 mph would take 40 hours
B @ 2 mph would take 20 hours
and C @ 8 mph would that 5 hours
to complete the trip each.

But after 5 hours C is at W and A is at 5 miles from R or 35 miles from W so has 35 miles (at that instant) to go back to get A
They will be travelling towards each other at 1 + 8 = 9 mph so will meet after 35/9 = 3.88 hours
At this instant A will be at 5 + 3.88 = 8.88 miles from R. or 31.12 miles from W. and B will be at 17 .76 miles from R or B is 22.24 miles from W
C picks up A and goes back to pick up B. It doesn't matter where C passes as long as he passes him C will pick up B and continue on to W
So C travels from pickup A to W (also with B picked up as he passes) at 8 mph so 31.12/8 = 3.89 hours and all arrive together
total time is 5 + 3.88 + 3.89 = 12.77 hours....................

[Guest]

" How much of last post is the question and how much is your working out. Is the illustration part of the question or to illustrate your understanding of the question. ' I didn't work out any part. The illustration was included with problem. The portion below that was a description of illustration.

" Are the first few lines up to "Calculate time of trip". the only bit that is the question " Yes, that part is the question.

I don't know how they arrived at 15.24 hrs.

[Guest]

Ok, I took what I reckoned was the simple easy option of what may have happened in reality as C picked up A and then B and continued on to W.

But I think it can be worked out for C dropping off A at G and then going back for B, then C + B only continue to W while A walks from G to W

It would be a good exercise to try yourself. Now that you have seen the reasoning applied in the earlier post.

A walks from G to W in same time as C goes from G back to meet B and then turns back to finish at W. C and B will arrive together just as A completes his walk from G to W

[Guest]

" It would be a good exercise to try yourself. Now that you have seen the reasoning applied in the earlier post. "

I don't know how.