- 1,631

- 4

....got it....

Thnx though!

Thnx though!

Last edited:

- Thread starter sutupidmath
- Start date

- 1,631

- 4

....got it....

Thnx though!

Thnx though!

Last edited:

- 287

- 0

x: person's x position

y: person's y position

X: raptor's x position

Y: raptor's y position

x = 5(sqrt2) (t+3)

y = 5(sqrt2) (t+3)

X = 12(sqrt2) t_x

Y = 12(sqrt2) t_y

If the raptor wants to catch you, t_x must equal t_y... because the person is on the line x=y and the raptor starts at the origin. So t_x equals t_y equals t.

Let T represent the time the raptor catches you (with t = 0 being when the raptor starts running). Then we must have that

5(sqrt2)(T+3) = 12(sqrt2)(T/2)

The equation is the same for y.

Solving for T...

5T + 15 = 6T

T = 15.

So it takes the raptor 15 seconds to catch the dude.

It can't very well be 1.5 s, can it? The raptor has only moved 22.5(sqrt2) = 32 total, while the person has moved 45 total. Even if the raptor was chasing the guy directly, he wouldn't catch him.

- 1,631

- 4

Comparing my solution with yours(the one that i did after my first post), i see that we have the same idea, in general terms with some slight differences, besides that you have a typo on your solution, because the speed of the raptor is 15sqrt{2}m/s and not 12sqrt{2}m/s so the final answer should be 6 s.Check this out.

X = 12(sqrt2) t_x

Y = 12(sqrt2) t_y

T = 15.

So it takes the raptor 15 seconds to catch the dude.

It can't very well be 1.5 s, can it? The raptor has only moved 22.5(sqrt2) = 32 total, while the person has moved 45 total. Even if the raptor was chasing the guy directly, he wouldn't catch him.

Cheers!

- 1,631

- 4

It looks like that's the case.What happened? Did the raptor eat your post instead of you??

Well, in fact, after i initially posted the problem and the 'solution', after a few hours i wanted to edit it since i found the correct solution, but while tring to post it the pf was having trouble or don't know what, so that's what was left from my post.

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