Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Sep 10th 2008, 07:03 PM   #1
Junior Member
 
Join Date: Sep 2008
Posts: 4
Question Constant Speed Problem

hey, i am gonna just write the problem just as it is in the textbook. I have thought about this one for a while, and yet have no idea where to start. I figured out the answer (I think) by guess and check, but i would really like to UNDERSTAND why the answer is what it is. Thanks, here it is:

Two runners approaching each other on a straight track have constant speeds of 4.50m/s and 3.50m/s, respectively, when they are 100m apart. How long will it take for the runners to meet, and at what position will they meet if they maintain these speeds?

It kinds looks like so:

Runner A 4.5m/s--->>>______________100m______________<<<----Runner B 3.5m/s
tcrocRE is offline   Reply With Quote
Old Sep 11th 2008, 04:07 AM   #2
Member
 
Mr Fantastic's Avatar
 
Join Date: Apr 2008
Location: Zeitgeist
Posts: 69
Originally Posted by tcrocRE View Post
hey, i am gonna just write the problem just as it is in the textbook. I have thought about this one for a while, and yet have no idea where to start. I figured out the answer (I think) by guess and check, but i would really like to UNDERSTAND why the answer is what it is. Thanks, here it is:

Two runners approaching each other on a straight track have constant speeds of 4.50m/s and 3.50m/s, respectively, when they are 100m apart. How long will it take for the runners to meet, and at what position will they meet if they maintain these speeds?

It kinds looks like so:

Runner A 4.5m/s--->>>______________100m______________<<<----Runner B 3.5m/s
Consider one of the runners standing still and the other approaching at 8 m/s. Calculate the time they meet.

Now consider both runners approaching with their given speeds and calculate the distance travelled by each in that time.
Mr Fantastic is offline   Reply With Quote
Old Sep 11th 2008, 07:06 AM   #3
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,152
Originally Posted by tcrocRE View Post
hey, i am gonna just write the problem just as it is in the textbook. I have thought about this one for a while, and yet have no idea where to start. I figured out the answer (I think) by guess and check, but i would really like to UNDERSTAND why the answer is what it is. Thanks, here it is:

Two runners approaching each other on a straight track have constant speeds of 4.50m/s and 3.50m/s, respectively, when they are 100m apart. How long will it take for the runners to meet, and at what position will they meet if they maintain these speeds?

It kinds looks like so:

Runner A 4.5m/s--->>>______________100m______________<<<----Runner B 3.5m/s
Another way. Say runner A starts at the origin and let x be the position that they meet at. Then we know
$\displaystyle x = 4.5t$
and
$\displaystyle 100 - x = 3.5t$

Solve the system of equations for t and x.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
constant, problem, speed



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
chair-lift at constant speed: confusion over passengers per minute, power, etc. furor celtica Kinematics and Dynamics 1 Aug 30th 2011 11:10 AM
speed problem christina Kinematics and Dynamics 1 Oct 12th 2009 03:41 AM
Reaction speed problem. ariol Kinematics and Dynamics 1 Jan 8th 2009 11:15 PM
Speed Problem dsptl Kinematics and Dynamics 1 Oct 17th 2008 03:16 AM
constant speed problem chocolatelover Kinematics and Dynamics 2 Sep 6th 2008 09:38 AM


Facebook Twitter Google+ RSS Feed