## silverxx one year ago kinamatics

1. silverxx

the dog from rest accelerates 5m/s^2 instantly at the sight of the boy and runs after a very scared boy. since the dog chases him, he accelerates 2 m/s^2 running from 10m/s seped. how far a distance the dog can overtakes the boy?

2. silverxx

ow thats speed not seped.

3. theEric

This one is not ambiguous, thankfully. It looks like it was translated from another language, though. Do you understand the situation?

4. silverxx

yes but no idea how to answer this one.

5. theEric

Actually, there is missing information. How far apart are the boy and the dog in the beginning? Without this, we can't find an exact answer. The answer would have a variable in it to account for this distance.

6. theEric

7. silverxx

yay. thats the only infos on the problem.

8. theEric

What I would do is come up with two equations: one to describe the position of the dog, one to describe the position of the boy. Then set them equal to each other, which makes the equation true for only the one time at which the dog catches the boy. When you find that time, you can plug it into the equation for either the dog's or boy's position equation to find out the position where they meet. Then you can find how far that is from where the dog started.

9. theEric

But time will depend on how far apart the dog is from the boy in the beginning. So you start of with your basic $$x=x_0+v_0t+\frac12at^2$$ equation. One for the dog, one for the boy.

10. theEric

So you'll plug in the values from the problem. Can you do that?

11. theEric

Well, $$x_0$$ for the boy will be unknown, so you let it be a variable. I think you'll be able to use the quadratic formula to get $$t$$. And then you can plug that $$t$$ into the equation for the dog's position. I have to go. Take care!

12. silverxx

thank you so much =)