A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
A car passes a landmark on a highway traveling at a constant rate of 55 kilometers per hour. One hour later, a second car passes the same landmark traveling in the same direction at 63 kilometers per hour. How much time after the second car passes the landmark will it overtake the first car?
Do not do any rounding.
anonymous
 one year ago
A car passes a landmark on a highway traveling at a constant rate of 55 kilometers per hour. One hour later, a second car passes the same landmark traveling in the same direction at 63 kilometers per hour. How much time after the second car passes the landmark will it overtake the first car? Do not do any rounding.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. First, think about what conditions mean that one car passes another...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well their distance travelled after the landmark must be same right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so then consider that the distance is x, alright? Then you know their speeds and the time difference. The second car starts one hour later. So, whatever time in which they reach at that same point, value of t for second car is t1 hrs. Now equate speed/time for both

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know how to do that. D:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, I really don't understand this. I need help.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to the attachment from Mathematica.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440487713058:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then, distance=speed*time, right? That would mean 55*t=63*(t1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, sorry but the question need t1 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I chose t to be time elapsed for first... so 6.875 will be the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I understand now! Thanks. I have 5 more questions like this to do and I think I can get them on my own now. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0shouldn't it be `t+1` since it was an hour later?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0`t1` would mean the second car left an hour before the first, no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I chose the other value for t for the question @Jhannybean. and no an hour late would mean lesser time elapsed right?.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0t represents time elapsed after crossing the landmark.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0time it takes for the first car to pass the landmark = t time it takes for the second car to pass the landmark an hour later = t+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let t be the time to catch the first car. 55 (1 + t) = 63 t // Solve t = 55/8 = 6.875

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for the medal.
Ask your own question
Sign UpFind 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.