How Far Does C Walk?Date: 08/06/2001 at 07:20:33 From: pollution Subject: Speed Three persons A, B, and C, travel from point X to point Y. They leave point X at the same time and follow the same route. A rides a motorcycle at a speed of 56 kilometres/h, B walks at a speed of 8 kilometres/h, and C walks at a speed of 7 kilometres/h. At the beginning, A gives a lift to B at X to travel a part of the journey and then comes back to pick up C. They all arrive at point Y at the same time. If B walks 2.8 kilometres, how far does C walk? Date: 08/07/2001 at 12:38:55 From: Doctor Greenie Subject: Re: Speed Hello -- Thanks for sending your question to us here at Dr. Math. Let's divide the problem into three time periods: t1 = time A is carrying B from X toward Y (C is walking) t2 = time A is heading back from Y towards X to pick up C (B and C are both walking) t3 = time A is again heading from X towards Y, carrying C (B is walking) The rates of A, B, and C in kph are 56, 8, and 7, respectively. Then we can write some expressions and equations using the given information: (1) B travels time t1 at 56kph and time (t2+t3) at 8kph; the distance he travels is 56t1 + 8(t2+t3) (1) (2) C travels time (t1+t2) at 7kph and time t3 at 56kph; the distance he travels is 7(t1+t2) + 56t3 (2) (3) B and C both travel the same distance, so (1) and (2) above are equal: 56t1 + 8(t2+t3) = 7(t1+t2) + 56t3 (3) (4) B walks 2.8 km: 8(t2+t3) = 2.8 (4) At this point we should stop and think about where we are going with this problem. The variables we have set up to solve the problem are t1, t2, and t3. However, there is no requirement that we find the values of those three variables. The question we are trying to answer is the distance C walks; this distance is 7(t1+t2). So, to answer the problem, we only need to find a value for (t1+t2) - we don't necessarily need to find the values for t1 and t2 individually. Now let's continue. From (4) above, we have 8(t2+t3) = 2.8 t2+t3 = 2.8/8 = 0.35 (5) Substituting (4) in (3) and simplifying, we have 56t1 + 8(t2+t3) = 7(t1+t2) + 56t3 56t1 + 2.8 = 7t1 + 7t2 + 56t3 49t1 + 2.8 = 7t2 + 56t3 (6) Now using (5) to rewrite the right-hand side of (6), we have 49t1 + 2.8 = 7t2 + 56t3 49t1 + 2.8 = 7t2 + 7t3 + 49t3 49t1 + 2.8 = 7(t2+t3) + 49t3 49t1 + 2.8 = 7(0.35) + 49t3 49t1 + 2.8 = 2.45 + 49t3 2.8 - 2.45 = 49(t3-t1) 0.35 = 49(t3-t1) 0.35/49 = t3-t1 (7) Now we can combine (5) and (7): t2+t3 = 0.35 t1-t3 = -0.35/49 t1+t2 = 0.35 -0.35/49 t1+t2 = 0.35(1-1/49) t1+t2 = 0.35(48/49) (8) We now have a numerical value for (t1+t2), so we can find the distance C walks: 7(t1+t2) = (7)(0.35)(48/49) = (7)(7/20)(48/49) = 12/5 = 2.4 km While we have answered the question which was asked, we can go further with the problem to learn more. If we let d be the distance between X and Y, then we can find that distance by equating the times that A and B traveled. B traveled (d-2.8) km at 56 kph and 2.8 km at 8 kph; the total time he traveled was (d-2.8)/56 + 2.8/8 A traveled (d-2.8) km with B in part 1 of his trip; he traveled back (d-5.2) km alone in part 2; and he traveled (d-2.4) km with C in part 3. Altogether, he traveled (d-2.8)+(d-5.2)+(d-2.4) = (3d-10.4) km, all at 56 kph; the total time he traveled was (3d-10.4)/56 Because A, B, and C all reached Y at the same time, the total travel time is the same for A and B: (d-2.8)/56 + 2.8/8 = (3d-10.4)/56 Solving this equation gives us the distance from X to Y as 13.6 km; once we have this information, we can determine the duration of each of the three parts of the trip and the distances each person traveled in each of those time periods. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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