Fact are facts, Watson !
"As per Inspector Lestrade, the accused house is on the bank of a river which is quarter mile wide and has current of 2 mph". said Dr. Watson
"Don't forget Watson, Just opposite his house, on the other bank, is victims house. The accused can swim across to his house." Sherlock Holmes added.
"But its impossible for him to reach there in less than ten minutes". exclaimed Dr. Watson.
"The fact are, the accused can swims at 2.5 mph in still water and his walking speed also is 2.5 mph."
"Can he do by a combination of swimming and walking?"
"Don't forget Watson, Just opposite his house, on the other bank, is victims house. The accused can swim across to his house." Sherlock Holmes added.
"But its impossible for him to reach there in less than ten minutes". exclaimed Dr. Watson.
"The fact are, the accused can swims at 2.5 mph in still water and his walking speed also is 2.5 mph."
"Can he do by a combination of swimming and walking?"
Labels: friday special, puzzle, thinktank
posted by Rajesh Lal at
8:02 AM
25 Comments:
It would take 6 minutes to swim across the river. That would put him .2 miles from the house. It would take 4.8 minutes to walk to the house so the total time would be 10.8 minutes. So by consideration it would take more than 10 minutes to make the trip with a combination of walking and swimming
Case 2 if the Accused walked upstream .2 miles and the clock started when he entered the water you can get across in 6 minutes to the house. But either way it would be just under 5 minutes to walk to that point.
I agree with Anonymous's calculations. But if the accused swims directly towards the opposite shore and doesn't let the current carry him downstream, I believe he could swim across in 7.68 minutes.
I calculate the hypoteneuse of a right angled triangle with one side .25 miles and one side .20 miles to be .32 miles. At 2.5 miles per hour, one could swim .32 miles in about 7.68 minutes.
I read the question and the only answer I can think of is, "What?!"
me thinks that if he walks far enough upstream and jumps in
his swim speed + current speed
should make the ans yes
Sorry Black Knight, you're sort of on the right track but your calculations are incorrect. You have mixed up velocity and distance vectors in the same calculation. The right angle triangle with sides of 2mph downstream and 2.5mph diagonally upstream (this is the hypotenuse)adds up to an effective velocity of exactly 1.5mph across the stream. Time to swim directly across= 0.25m/1.5mph *60mins= exactly 10 minutes. So the accused can make it, if he heads upstream at an angle of about 59 degrees.
A little trigonometry applied to mechanics make this problem doable.
Shortest distance between two points being a straight line means that the quickest way to cross the river will be stright across, swimming at an angle to compensate for the current of the river.
We can calculate the angle required by saying the at componant of the swimming speed needed to counteract the current river is 2.0mph
So we can say that 2.5 sinx = 2.0
Solving gives x = inverse sin (0.8)
The resulting horrizontal speed is then given by 2.5 cos x
Using x from the first part gives a horizontal speed of 2.5 x 0.6 = 1.5mph.
At 1.5 mph he will travel 0.25 miles in 10 minutes.
So if the suspect swam at exactly 53.13 degrees he would have reached the opposite side in 10 minutes, and would have been able to commit the crime.
|\
2.0mph | \ 2.5mph
current |__\ swimspeed
1.5mph
resultant horizontal speed
now what if the wind is blowing? It can change current direction which in turn could decrease the time across as well ?
Seeing as this is Sherlock Holmes and nothing is ever what i seems ill say there was a bridge he walked across
either it was so shallow he was just walking in water
there was a bridge
or he grabbed onto shamu
the water is so shallow he walked
he walked on a bridge
he garabbed on shamu
yup agree with Sir. but Sir's calculations only prove that it is possible to swim in 10mins.
can you find the optimum angle in which to swim (or can you prove that 53.13 deg is the optimum angle) so that swimming time is least? since an angle with more horizontal speed will also mean a longer distance..sorry i don't have an answer cus haven't given that much thought.
just taking that you can only swim..of course you can also include walking then there'll just be a change in absolute numbers. sorry i didn't check any numbers..just scanning the methods...i think there's definitely more people here who are right but sir's method was the most easy to follow...
wouldn't that depend on how deep the water is as to see if you can walk across it?
This post has been removed by the author.
The solution of swimming angles ranges from an angle of 8.8° to 53.1°. The latter is a solution found by multiple users before me. However in contradiction to their claims this is not the fastest way to get accross the river. In fact this is only the bear minimum, reaching the house of the victim in just ten minutes. The fastest way to get there is to swim at an angle of 33.7°. At this angle you can reach the victim in just under 9 minutes.
Time to cross the river (T1):
T1=1/4 mile / cos(alpha) / 2.5 mph
=1/(10*cos(alpha))
(alpha is the angle of the swimming direction(upstream, excluding stream) across the river with respect to the normal)
the drift (X) is than:
X=T1 * 2 mph - tan(alpha)*1/4
walking back this drift over the shore takes T2:
T2= X/2.5 mph
the sum of the times amounts to T:
T=T1+T2=T1*(2/2.5+1)-tan(alpha)/10=4.5/(25*cos(alpha))-tan(alpha)/10
T<1/6 --> 8.8°<=alpha<=53.1°
Sir's method is right in the sense that it gives the shortest distance (through the water). However it is not the fastest way to get to the victim because the effective speed through the water amounts only to 1.5. where one could go as fast as 2.5 over the land.
me thinks
as i`ve said before yes
change 1/4 mile to 440yds
5 mins upstream /\ walk at 2.5 mph
converts to 366.6 yds
sq 0f 440 + sq of 366.6 = sq of hyp
which ended up being 572.7 yds
(thats result sq inverted)
5 mins left so 2.5 swim speed
with correct angle + current speed =2.5 + 2 = 4.5 with 5 mins
converts to
660 yds > 572.7 hyp
thats 88 yds diff
allowing for other minor factors should still make it
No matter how you approach this question, the answer is inevitably no. It's very possible for the criminal to have returned to his house in 10 minutes. a rather elementary physics problem.
Say we take a farfetched example, that he just decided to swim across the river, allow himself to be carried and wherever he ended up, he walked back to his house from there. This means that he did not fight against the current, rather just focus on swimming 2.5 mph towards the shore and let the current displace him in either direction of the river.
If this is the case, this means that he'll still be swimming towards his shore at 2.5 mph, but will be carried some length down the river, the distance that he can travel in 2 mph in the time it takes to cross .25m at 2.5 mph.
This is assuming the the acceleration of him by the river is instantaneous and he swims at a constant speed of 2.5mph. This will realistically reduce a certain amount of time from his total trip because it makes him go faster than he actually will in real life.
It will take him 6 minutes to swim across the river. While being displaced .2 miles in either direction. Then walking at 2.5 mph it'll take him 4.8 minutes to walk .2 miles back to his house. That's a total of 10.8 minutes and it doesn't include the time he needs to get out of the river, or his initial velocity of zero while getting into the river.
I'm currently working through the solution of if he manages to swim against the current so that he's not displaced by the river current and lands at his house, but I think it will be much longer than 10 minutes.
Got it. In order to swim directly across the river at a constant velocity, the person must swim at 53.1 degrees in the opposite direction of the current. You can see this if you draw a current of 2 mph going right (any length of a line). At the right end of the line, draw a straight line down to symbolize crossing the river, and from that point at the bottom of the vertical line, draw a slanted line towards the left end of the top horizontal, 2mph line. You'll get a right angle triangle with the 90 degrees angle at the top. The slanted line is the 2.5 mph vector while the top horizontal is the 2 mph to the right. This symbolizes that he's swimming to the left while the river pushes him to the right.
This way, you can add the vectors since you assume velocity is constant. Using opp/hyp=sin(theta), we see that in order to achieve the velocity (the line directly up and down) of directly across the river, the angle theta (between the 2.5mph line and the vertical line) must be 53.1 degrees. Then using adj/hyp=cos(theta), adj being the vertical line, we can find out that the resultant vector is 1.5 mph upwards. This translates into 0.025 mpm (miles per minute) which is exactly 10 minutes to cross the river. However, again this doesn't fact in gradual acceleration, which would've increased the time and getting out of the river, which increases it further. Either way, I believe that it will have taken the person longer than 10 minutes, no matter what.
Ben, your persuming that swimming straight across the river (withstanding the current) and no walking is the fastest way to get to the house, which as Bas already pointed out is wrong. A combination of swimming slightly against the current and walking can be faster.
If µ is the angle the acussed was to swim against the current, heres how long it would take (in seconds):
T(µ) = 648/cos µ - 360 tan µ
This is equivalent to what Bas worked out. You'll notice that if you choose µ = arcsin 0.8 (~53.1°), you get the 600 sec (10 min) you came up with. But that's not the optimum.
To calculate the optimal angle, we need to find out where the derivative becomes 0.
T'(µ) = -sin µ (-648/cos²µ) - 360/cos²µ
= (648 sin µ - 360)/cos²µ = 0
=> µ = arcsin(360/648) = ~33.7°
(just like Bas says)
Swimming at this angle, the acussed could have crossed the river in 7 min 13.0 sec, climbing ashore .074 miles dowmsteam from the house, then walked that distance in 1 min 45.8 sec, reaching the house in 8 min 58.8 sec.
To the anonymous guy before Ben: that doesn't work. To reach a velocity of 4.5 mph, the acussed would have to swim straight downstream, you can't use the current to get across the river faster.
I feel rather silly for not having read that! Thanks for pointing that out... that is faster!
to mo
554yds(corrected hyp)
i stand corrected if not 4.5 then somethink close .
decided to have a go
i came up with 3.98 x axis (abt 3mins)
and 3.52 y axis which means using current u can cross faster.
correction (4.8 mins )
Anonymous:
If the accused walks 5 min upstream, he will have walked 366.7 yards. The river is a quater of a mile (440 yards) wide, so his speed in the water will be nowhere near 4.5 mph, it will be more like 3.2 mph (which is still faster than swimming without a current).
In any case, the fastest way to get to the other river bank, never mind the drift, is to swim straight towards it, which will take 6 minutes. No, you can't use a sidewards current to beat that, you can't cheat physics.
me thinks
hyp = 572.7 yds (sq of 440 + sq of 366.66 = hyp(inv))
you are 366.66 yds(x axis) upstream (5 mins walk at 2.5 mph)
swiming at 2.5 mph at a angle
y ( 440(y axis) ) / h (572.7 (hyp)) = 0.7682 +
= sin(inv) = 50.2
with the current(constant = 2mph)
calulate vector(50.2 and 2.5+2) speed
along x axis (along the vector in the direction of x axis(366.6 yds))
50.2 cos times 4.5mph = 2.88
convert to yds per min
=84 .48
calulate vector(50.2 and 2.5+2) speed
along y axis (along the vector in the direction of y axis (440 yds))
50.2 sin times 4.5 = 3.45
convert to yds per min
=101.4
along x axis
366.6/84.48= 4.34 mins
along y axis
440/101.4 = 4.34 mins
4.34+5=9.34
so less than 10 mins
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