20 replies to this topic

[ujjagrawal]

There is a square ABCD with a Cat at A and a mouse at B. The mouse starts walking towards C, while the Cat walks directly towards the mouse. If the Cat walks n times as fast as the mouse, and catches the mouse at C, what is the value of n?

Additionally can you also find what is the equation of the Cat's path?

[hhh3]

wel.. this is too mathematical for me.. any way i'm gonna tak a stab n the dark...

Spoiler for hmmm...

the value of n is 2/pi ??? by any chance?? hi hi...

[bonanova]

Spoiler for First off

The cat clearly travels faster than the mouse.
The mouse's path is a side of the square.
The cat's path length is longer than a diagonal but shorter than two sides.
So n lies between 1.414 and 2.

Spoiler for looks like

The slope seems to be somewhat proportional to the path length.
So I'm guessing a catenary. In which case,

n = 1.6161 times as fast as the mouse.

If the cat starts at the origin and the mouse goes from [1,0] to [1,1]
and if the curve is a catenary, the cat's path is given by

y = [cosh(nx)-1]/n

[ndk]

Spoiler for please use spoilers

My guess would be that the cat must follow an arc covering a quarter of the circumference of the circle with the radius equal to the length of one side of the square ABCD. Therefore speed would be 1.5708 times that of the mouse.

[ujjagrawal]

No proper solution... apart from few guesses... Still waiting for anyone to solve this one ?

[phil1882]

Spoiler for hmmm

so clearly the mouse will run in a straight line.
lets say the square is 1 meter and the mouse runs at the speed of 1 meter per sec.
mouse = 1 m/s
the position of the mouse after t seconds has passed would be
mouse = t meters
now the cat starts at 1 meter straight away from the mouse, perpendicular to the direction the mouse is running.
we know the cat catches the mouse after 1 sec.
this suggests to me that after a 1/2 second, the cat is half way to catching the mouse. after 7/8 a second, the cat is 7/8 of the way to catching the mouse. and so on.
so based on this, its sounds like the equation for the motion of the cat is
cat = n*t meters
so velocity of the cat is sqrt(2).

[fabpig]

Spoiler for Without the maths

I agree with Bonanova. I think the apex of the catenary lies at a point on the BD diagonal and the intersection of a line drawn between the midpoints of AB and BC

btw. I don't think negative reputation points are intended for wrong answers: more for rude or insulting statements.

[curr3nt]

Spoiler for How about...

n = pi * (2(AB² + BC²))^(1/2) / 4BC

It looks like the cat travels along a quarter of the circumference of a circle/oval. So if AB = 1 and BC = 1 the cat travels pi/2 to the mouse's 1.

[Yoruichi-san]

Spoiler for The maths

Defining my coordinate system (and orientating the cat the mouse) so that mouse starts at (s,s), wgere s=length of one side, cat starts at (0,s). Also defining mouse speed = v, and X(t) to be my function of the x position of the cat at time t. At time t, the mouse's distance from its starting point is vt. The cat's distance from it's starting point is X, the velocity of the cat is nv. Hence the distance b/w the cat and the mouse will be sqrt((s-X)²-v²t²).

Then the speed of the cat, in the x-direction, dX/dt = nv²t/sqrt((s-X)²-v²t²)

To be honest, it's been a while since diffEQ, so I don't recognize the equation or way to solve it off the top of my head. Bonanova and fabpig are probably right about it being a catenary. But solve this differential equation, and you will have the function (you can put it in terms of y, if you want, instead of t, by using the relation b/w the vt and y). You know it will take t_f=s/v for the mouse to reach it's destination, so plug that in, put the other side equal to s (the total X-length the cat travels in that time), and solve for n.

[curr3nt]

Spoiler for drawing...

In order to draw it out I had to make the cat move towards where the mouse was going to be. Since if the cat only moves to where the mouse was it couldn't catch it.

Here is where the cat moves in 1 - 4 and 6 steps.

This is what lead me to the oval/circle path.