1) Jim, Lydia, and Isabel are visiting China. Unfortunately they are stuckin a s
ID: 3099793 • Letter: 1
Question
1) Jim, Lydia, and Isabel are visiting China. Unfortunately they are stuckin a seemingly infinite traffic jam. The cars are moving at a very slow (but constant) rate. Jim and Lydia are 25 miles behind Isabel. Jim wants to send a sandwich to Isabel. So he hops on his motorcycle and rides through traffic to Isabel, gives her the sandwich, and rides back to Lydia at a constant speed. When he returns to Lydia, she has moved all the way to where Isabel was when Jim started. In total, how far did Jim travel on his motorcycle?
(a) Before any computations are done, use common sense to guess thesolution to this problem.
(b) Choose numbers for the above problem and solve the problem withthese numbers. Use this computation to test your guess and possibly make a new guess.
(c) Use algebra to solve the problem.
*** PLEASE SHOW ALL WORK, AND USE THE EQUATION EDITOR PLEASE. WILL RATE LIFESAVER IF GOOD!
Explanation / Answer
Lydia and Isabel will be simply L and I, and Jim will be J.
At the beginning, J & L are in point A, and I is in B, 25 miles ahead.
So AB = 25
They (L & I) are moving formard at speed V (the same speed because of the traffic jam), while J does the trip at a faster speed W (W>V), catches up to I and gets back to L. By then L is in point B, so she has travelled 25 miles, and therefore so has I who must then be in point C, another 25 miles ahead of B.
So, while L & I have done 25 miles at speed V, J has done 25 + d to catch up with I, and then did the same d distance back to rejoin L. By then I is another 25 miles ahead in point D:
------A------------------------B----------------D-------C----------------------->
AB=BC=25
J did the distance AD at speed W. The time t required for this trip is:
t = AD/W
During that same time t, I had gone from B to D, going at speed V, so we also have:
t = BD/V
Then J comes back to meet L, and they meet in B.
This return takes a time “s”, which J does at speed W so:
s = BD/W
While I continues to point C at speed V so:
s = DC/V
During all the time, (s+t), L has moved from A to B at speed V so:
(s+t) = AB/V
So if we call the distance AB “b”, we also have BC=b, and the distance BD is “d”, and CD is BC – BD = b-d.
(a) Guessing …
Now intuitively based on all that one would guess D is somewhere past the middle of BC. This is because it surely took longer to catch up with I, going in the same direction and only benefitting from the speed difference, than it took to get back to B and meet L , coming towards each other at the total of the two speeds. So one guess would be that distance d is between 2/3 and 3/4 of the distance b (25 miles). So if we ballpark d = 0.7 times b or around 17.5 miles - That would mean J has travelled the distance b (25 miles), then two times distance d or 35 miles, so in total it would look like he did approximately 60 miles or about that.
(b) Let us take a sample value for time “t”, say 1 hour:
The cars are then moving at 25 miles/hour. In one hour the motorcycle does 25 miles plus distance d TWICE, say 25+2d at speed W, while the car does 25 miles. So speed W is:
W = (25+2d) per hour.
The return from point D to B or distance d is done at speed W by the motorcycle in the same time the second car goes D to C (distance b-d) at speed V. So the speed ration W/V is equal to d/(b-d) so that gives us:
W = dV/(b-d) = 25d/(25-d)
So we have two expressions for W and we put them together:
W = (25+2d) = 25d/(25-d) this implies: 25d = (25+2d)(25-d) or:
625 +50d -25d - 2d2 = 25d ==> simplifies as 625 – 2d2+ 25d = 25d ==> 2d2= 625
Or d is the square root of 625/2 which is about 17.7 miles
And that means Jim travelled 25 + 17.7 + 17.7 = approx 60.4 miles.
(c) Let us lay out the equations and solve that pure algebra:
(1) t = AD/W OR t = (b+d)/W OR t.W = b+d
(2) t = BD/V OR t = d/V OR t.V = d
(3) s = BD/W OR s = d/W OR s.W = d
(4) s = DC/V OR s = (b-d)/V OR s.V = b-d
(5) s+t = AB/V OR s+t = b/V OR s.V +t.V = b
Notice equations (2) and (4) added together become (5): s.V + t.V = b-d+d = b, so it would seem wehave consistency, but also we have only 4 equations (5 is useless as it is dependent on 2 and 4), and we have 5 variables: s,t,V,W,d…
The interesting equations are pairs:
Take (1) and (2) and DIVIDE (1) by (2) you get:
(1)/(2) => t.W / t.V = (b+d)/d => or simplifying the “t” out we get equation (6):
(6) W/V = (b+d)/d
Now we do the same with (3) and (4) and we get:
(3)/(4) => s.W / s.V = d / (b-d) => or simplifying the “s” out we get equation (7):
(7) W/V = d/(b-d)
Now equations (6) and (7) together give:
(6)=(7): W/V = (b+d)/d and (7) W/V = d/(b-d) ===SO===> (8) (b+d)/d = d/(b-d)
By multiplying both sides by d first and then by (b-d) we get:
(8) ==> d(b+d)/d = dd/(b-d)
(8) ==> d(b+d)(b-d)/d = dd(b-d)/(b-d)
(8) ==> (b+d)(b-d) = d2
(8) ==> b2-d2 = d2
Finally this gives 2 d2 = b2 ==> gives d2 = b2 / 2 ==> d = b/2
If b= 25 that gives d= 25/1.414= 17.67 miles
Then based on equations (2) and (3), we can say:
(1) t.W = b+d, and (3) s.W = d => s/t = d/(b+d) OR s= dt/(b+d)
So s= 0.414 t
And going back to (6) W/V = (b+d)/d
We get W = 2.414 V
If t = 1 hour
Then s = approximately 25 minutes,
V = 25 miles/hour and W = 60.35 miles / hour
The logic is surely good, but check the numeric calculations with your calculator…
If you have questions email raphhome@xplornet.com
RAPH
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.