A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

silverxx
 9 months ago
Best ResponseYou've already chosen the best response.0the 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?

silverxx
 9 months ago
Best ResponseYou've already chosen the best response.0ow thats speed not seped.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1This one is not ambiguous, thankfully. It looks like it was translated from another language, though. Do you understand the situation?

silverxx
 9 months ago
Best ResponseYou've already chosen the best response.0yes but no idea how to answer this one.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1Actually, 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.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1Is there any more information in that problem?

silverxx
 9 months ago
Best ResponseYou've already chosen the best response.0yay. thats the only infos on the problem.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1What 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.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1But 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.

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1So you'll plug in the values from the problem. Can you do that?

theEric
 9 months ago
Best ResponseYou've already chosen the best response.1Well, \(x_0\) 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!
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.