Find the distance at which the peacock will catch the snake

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Homework Statement



A peacock perched on top of a 12m high tree spots a snake moving towards its hole at the base of the tree from a distance equal to thrice the height of the tree. The peacock flies towards the snake in a straight line and they move at the same speed. At what distance from the base of the tree will the peacock catch the snake?

Homework Equations



1)equation of line
2) Pythagoras theorem

The Attempt at a Solution



I used the equation of the line and tried to find the slope

m= [itex]\frac{y-yo}{x-xo}[/itex]

[itex]\Rightarrow[/itex] m=[itex]\frac{0-12}{36-0}[/itex]

[itex]\Rightarrow[/itex] m=[itex]\frac{-12}{36}[/itex] = -0.33


and then tried substituting this slope in the other equation of line in which the peacock catches the snake.

But I think this is wrong

Please give me hints on how to go about this problem.
 
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mfb said:
That slope would require a stationary snake.

Did you draw a sketch? This will help a lot.

yes, I have drawn the sketch.
and I know that this problem is very easy, but I can't figure out how to go about the solution.

Please give me some hints.
 
Mark the position where they hit each other, and define some variable expressing its position. Calculate the distance both animals travel, and use that they have to be equal. This allows to get the value of your variiable.
 
mfb said:
Mark the position where they hit each other, and define some variable expressing its position. Calculate the distance both animals travel, and use that they have to be equal. This allows to get the value of your variiable.

the distance are equal
[itex]\Rightarrow[/itex] 36-S1=12-P1
[itex]\Rightarrow[/itex] S1-P1=24

Is this correct?
If yes, then how do I get the second equation?
Please help...
 

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P1 is the y-coordinate where the peacock catches the snake,
in this case P1=0

therefore

36-S1=12-0
S1=24

But on the book they have given the answer as 16 meters.
 
Since the time taken to reach the point by both and their speeds are the same, just equate the distances that they have traveled.
 
Sunil Simha said:
Since the time taken to reach the point by both and their speeds are the same, just equate the distances that they have traveled.

I did that in my previous post and I got the answer 24 meters, but the correct answer is required to be 16 meters.
 
gnrlies00 said:
I did that in my previous post and I got the answer 24 meters, but the correct answer is required to be 16 meters.

the answer does come to be 16 m the correct equation is
[itex]\sqrt{12^2 + s_1^2} = 36-s_1[/itex]