- #1
brotherbobby
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- Homework Statement
- Mail is carried between two landing stages M and K by two launches. At the appointed time the launches leave the landing stages, meet each other, exchange the mail and return. If the launches leave their respective launching stages at the same time, the launch departing from M spends 3 hours both ways and that from K 1.5 ##-## hours. The speeds of both launches are the same relative to water. Determine ##\underline{\text{graphically}}## how much later should the launch depart from M after the other one leaves K for the two to travel the ##\underline{\text{same time}}##. (my emphasis)
- Relevant Equations
- If the speed of river is some ##v## and the speed of launch relative to river ##u,## then motion downstream is ##u+v## and upstream is ##u-v.\quad (u>v)##
As it happened, I could solve the problem analytically but not graphically. I describe both methods below.
##(\rm{I})-\text{Analytical approach}##
The problem is divided into two parts. In the first part, both launches leave their stations at the same time.
I draw the situation of the problem alongside. Let the stations be a distance ##d## apart and the river to flow with a speed ##v##. Let each launch move with speed ##u## relative to river. Let the launching station M also be te origin, O. Both launches meet at a distance of ##x## (from O).
Launch 1 leaves M and 2 leaves K - since the launch from M takes longer to go and return, 1 it must be travelling downstream first for a greater distance also to meet 2. Let us use single primes ##'##s for forward motion and double primes ##''##s for return.
Time taken by both launches are the same for them to meet : ##\small{t'_1=\dfrac{x}{u+v}=\dfrac{d-x}{u-v}=t'_2\Rightarrow x=\dfrac{u+v}{u}\dfrac{d}{2}=x_1\quad \left(>\dfrac{d}{2}\right)}##. The distance travelled by launch 2 ##\small{x_2=d\left( 1- \dfrac{u+v}{2u} \right)=\dfrac{u-v}{u}\dfrac{d}{2}}##. The common time for them both ##t'_1=t'_2=\dfrac{d}{2u}##. To return, the first launch will take a time of ##\small{t''_1=\dfrac{u+v}{u-v}\dfrac{d}{2u}}## and the second launch ##\small{t''_2=\dfrac{u-v}{u+v}\dfrac{d}{2u}}##. Adding the forward and return times for both launches and equating them to the times given, we find the total time for launch 1 ##t_1=\dfrac{d}{u-v}=3\quad\color{blue}{(1)}\quad## and for launch 2 ##\quad t_2= \dfrac{d}{u+v}=\dfrac{3}{2} (=1.5)\quad\color{blue}{(2)}##
Launch 2 here travels for a time ##\Delta t## and is at a position ##(x_2)_{\Delta t}=d-(u-v)\Delta t## and arrives at N. If they both meet at P at a distance ##x## (from O or M), then the time taken by 1 ##\small{t'_1=\dfrac{x}{u+v}}##. This the same as the time taken by 2 to travel the remaining distance to P, ##\small{t'_2=\dfrac{d-(u-v)\Delta t-x}{u-v}}## which, after some algebra, simplifies to ##x=\dfrac{u+v}{2u}\{d-(u-v)\Delta t\}=x_1##. The distance covered by launch 2 ##x_2=d-x=d-\dfrac{u+v}{2u}\{d-(u-v)\Delta t\}=\dfrac{u-v}{2u}\{d+(u+v)\Delta t\}##, after some algebra.
The time for 1 forwards to meet : ##\small{t'_1=\dfrac{x_1}{u+v}=\dfrac{d-(u-v)\Delta t}{2u}}\quad (3)##. Likewise, the time for launch 2 forwards to meet 1 ##\small{t'_2=\dfrac{x_2}{u-v}=\dfrac{d+(u+v)\Delta t}{2u}}\quad (4)##.
Note that these times are different despite a "meeting" taking place., because launch 2 travelled for a longer time ##\Delta t##.
The time for 1 to return to his station M : ##\small{t''_1=\dfrac{x_1}{u-v}=\dfrac{u+v}{2u(u-v)}\{d-(u-v)\Delta t\}}\quad (5)##. The total travel time for launch 1 can be found by adding ##(3)+(5)##, which leads to ##\small{t_1=t'_1+t''_1=\dfrac{d-(u-v)\Delta t}{u-v}}\quad (6)##.
The time for launch 2 to return to his station K : ##\small{t''_2=\dfrac{x_2}{u+v}=\dfrac{u-v}{2u(u+v}\{d+(u+v)\Delta t\}}\quad (7)##. Hence the total travel time for launch 2 can be found by adding ##(4)+(7)\Rightarrow\small{t_2=t'_2+t''_2=\dfrac{d+(u+v)\Delta t}{u+v}}\quad (8)##.
But the two times are given to be equal, ##t_1=t_2\Rightarrow\small{\dfrac{d-(u-v)\Delta t}{u-v}=\dfrac{d+(u+v)\Delta t}{u+v}\Rightarrow \dfrac{d}{u-v}-\dfrac{d}{u+v}=2\Delta t\Rightarrow \Delta t=\dfrac{vd}{u^2-v^2}}\quad (9)##.
From ##(1), (2)## above, ##\small{\dfrac{d}{u-v}-\dfrac{d}{u+v}=3-\dfrac{3}{2}\Rightarrow \dfrac{2vd}{u^2-v^2}=\dfrac{3}{2}\Rightarrow \dfrac{vd}{u^2-v^2}=\dfrac{3}{4}}##. But from ##(9)## this is the value of ##\Delta t##. Hence ##\Delta t = \dfrac{3}{4}\,\text{hr}\Rightarrow\boxed{\Delta t= 45\,\text{min}}\quad{\color{green}{\Large\checkmark}}##
This answer matches with the text.
##(\rm{II})-\text{Graphical approach}##
As before, there are two cases - when the launches leave simultaneously and when launch 1 leaves M after a delay of time ##\Delta t.##
Launches 1 and 2 leave stations M and K shown in red. They meet at even P and return to their stations. We are given launch 1 takes a total of 3 hours, so MR = 3, while launch 2 takes 1.5 hours, so KQ = 1.5. The triangles KPQ and RPM are similar. Since base MR is twice KQ, the distance to P (and back) must be twice by M as for K. Hence we have ##x_1=2x_2##. Thus 1's speed must be twice 2's speed : ##u+v=2(u-v)\Rightarrow u=3v##, where ##u,v## are the speeds of the launches and river respectively.
##(\rm{b}) -\text{1 leaving M after a time }\Delta t##
Here launch 1 leaves M after a delay, they meet at P and return to their landing stations. In doing so, it's given that they take the same times, so KQ = NR. Hence triangles KPQ and RPM are congruent, implying that distances ##x_1=x_2##, so they both meet midway between the landings.
I could not glean further information from the graphs in order to answer the question : what is ##\Delta t=?##
Request : A hint as to how to go about solving the problem graphically.
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