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Speed/distance/time problem
By Ellie Tehrani (P1358) on Sunday, January 23, 2000 - 12:02 pm:

Two people, A and B are at a distance s from each other. A is next to a dog D. A and B start walking towards each other, both with speed u. At the same time D starts running toward B with speed v where v>u. When meeting B, D turns around immediately and runs back to A and so on. Assume that D's speed is v all the time and that D never stops when running between A and B.

In the easiest possible way find:

i) After how long time (expressed in terms of u, v, s) A and B will meet.

ii) How long distance will D have run then?

iii) Determine the ratio u/v in order for D to have run twice the distance of either A or B.

Appreciate an answer before 25/1.
Thank you in advance.
Yours
Ellie

By Dan Goodman (Dfmg2) on Sunday, January 23, 2000 - 02:26 pm:

This sounds like a homework problem, so I'll just give you a hint rather than a solution. Firstly, the question is the same if B is standing still and A walks with speed 2u (can you see why this is true?) Secondly, the dog is running at the same speed the whole time until A and B meet, this is important. Once you have these you should be able to solve the third part by translating it into an equation and solving it.

By Anonymous on Monday, January 24, 2000 - 06:35 pm:

i) by the formula, speed=distance/time, rearranged to time=distance/speed, the time is s/2u because A has to walk distance s/2 at speed u

ii) rearranging again we have distance=speed×time, the time is s/2u as above and the speed is v, so the distance is vs/2u

iii) since D and A have been moving for the same length of time at constant speeds, in order for their distances to be in ratio 1:2 we simply need their speeds to be in ratio 1:2, ie u/v=1/2 (since we know that v>u it must be 1/2 and not 2)