# Doubt : Pokemon Hunter and the Rogue Brook [closed]

I was working my way through some Puzzles in Discrete Maths by Rosen, when I came across the following question:

• A Pokemon Hunter is rowing upstream a brook
• As he passes under the 'bridge-of-curse', he throws a Pikachu into the brook
• 5 minutes later, he realises that Pikachu will die and he should not have done that
He rows back and picks Pikachu 3km. downstream of the 'bridge-of-curse'

What is the speed of the river flow?

- Distance : 3km
- Time : 5 minutes + 5 minutes ( due to the stream's frame of reference ) = 10 minutes
- Speed = 3000 / 600 m/s = 5 m/s

Doubt:

Am I correct ? Seems a bit too easy ...

• I'm wondering where you get the 10 minutes from? – Lyrion Jan 12 '15 at 10:09
• Poor Pikachu! So mean. – Golden Dragon Jan 12 '15 at 14:27
• All the answers assume the Pikachu will simply float downstream. They can swim rather well, so I say "not enough information given". – Set Big O Jan 12 '15 at 14:59
• This question appears to be off-topic because it is a textbook-ish math problem that belongs on math.SE. – xnor Jan 13 '15 at 23:07
• You did not provide enough justification to show how you are using a moving reference frame and make several assumptions but your calculation is correct. – kaine Jan 16 '15 at 14:03

You assume it takes him 5 minutes back as well. Here is a sketched solution:

s = speed of stream
r = speed of rowing in water with no current
t = time since drop to pickup


We have:

s * t = 3 // Pikachu travelled 3 kilometers for that time
t = 3 / s


We also have:

(t - 5) * (r + s) = 5 * (r - s) + 3


Which means the distance travelled returning is the distance going away + 3 kilometers. From here on we have:

(3 / s - 5) * (r + s) = 5 * (r - s) + 3
3r/s - 5r + 3 - 5s = 5r - 5s + 3
3r/s - 5r = 5r
10r = 3r/s
10rs = 3r
10s = 3


s = 10/3 = 3.3333 kilometres per minute

s = 3/10 = 0.3 kilometres per minute


This seems quite fast to me, if you see a typo in my calculations, feel free to edit or comment. Note $t =10$ which matches your assumption.

• He's rowing upstream originally, so his speed then is r - s. In the second part his speed is (r + s) and he covers the original distance plus three more km. So it should be $(t - 5) * (r - s) + 3 = 5 * (r + s)$, which you will find resolves to s = 3/10 = 3km per minute or 5m/s. – Callidus Jan 12 '15 at 13:15
• @Callidus t - 5 means the time he takes to come back. He's rowing downstream during that time. 5 is the time he rows upstream. My only mistake was in the last equation :) – dmg Jan 12 '15 at 13:28
• Right you are - I just reversed your entire equation, got the right answer, and missed the typo you spotted in your last line. Gah. – Callidus Jan 12 '15 at 13:33
• Your answer starts with "you are incorrect" but proceeds to exactly the same solution? Only in a more complicated way? – Falco Jan 13 '15 at 8:37

The original answer is correct but not explained. Viewed from the river's frame of reference, the Pikachu stays still; so the Hunter has simply rowed away for some period of time, and then rowed back in 5 minutes.

The Hunter is rowing at a constant speed in the river's frame of reference, but in the opposite direction in each time period. The Pikachu has stayed still, so the distance out and back must be the same, and the times must also.

That's 10 minutes total, and in that time the river has moved 3km in the bridge's frame of reference. So the river's speed relative to the bridge is 3km / 10 minute.

• Exactly what I wanted to write, with changed frame of reference the river stays still and just the landscape flows along 3km in 10 mins – Falco Jan 13 '15 at 8:39

However you might need to justify why 10 minutes is the correct time.

This can be done as follows;

The boat will travel at a speed $v_b=v_m+v_c$, ie. the total velocity is the sum of velocity from muscle plus velocity from current. $v_c$ is constant, while $v_m$ changes direction when the hunter turns.

The pokemon travels at speed $v_p=v_c$.

Distance is speed times time. $d_p=v_c*t=3km=d_b=(v_m+v_c)*t_{up}+(-v_m+v_c)*t_{down}=v_m*t_{up}-v_m*t_{down}+v_c*t$

Hence $v_m*t_{up}-v_m*t_{down}=0$ and $t_{up}=t_{down}$

Or stated in a simpler manner; the displacement due to current is the same for boat and pokemon, so the timing will be the same as if there was no current.

This assumes that the current is constant and that the only drag is from current, without those assumption the question lacks sufficient information to be solved.

You cannot directly assume that the total time taken is 10 min.

Assume,
speed of boat = b
speed of river = r

Let the total time taken be t
Since pikachu has travelled 3km in t time,
speed of river
r = 3/t
rt = 3

Now, suppose the boat has travelled a distance of x upstream from the 'bridge-of-curse' in 5min.
x = (b - r) * 5

Also, while coming downstream, the total distance covered is (x + 3),
while the speed will be (b + r) and time taken will be (t - 5)
So,
x + 3 = (b + r) * (t - 5)

(b - r) * 5 + 3 = (b + r) * (t - 5)
5b - 5r + 3 = bt + rt - 5b -5r
10b + 3 = bt + rt
We already have rt = 3
10b + 3 = bt + 3
10b = bt
t = 10 min

Speed of river, r = 3/t = 3/10 = 0.3 km/min