How to find Displacement

S

staceyrho

Junior Member
Joined
Jan 12, 2007
Messages
76
I have a problem, and I need to find the displacement of a runner. Here is the question:
Runner A is initially 6.24 mi W of a flagpole and is running with a constant velocigy of 4.06 mi/h dye east. Runner B is intitally 3.43 mi E of the flagpole abd us running with a constand velocity of 4.99 mi/h due W. Consider East to be the positive direction.
What is the displacement of runner B from the flagpole when their paths cross? Answer in units of mi.
 
staceyrho said:
I have a problem, and I need to find the displacement of a runner. Here is the question:
Runner A is initially 6.24 mi W of a flagpole and is running with a constant velocigy of 4.06 mi/h dye east. Runner B is intitally 3.43 mi E of the flagpole abd us running with a constand velocity of 4.99 mi/h due W. Consider East to be the positive direction.
What is the displacement of runner B from the flagpole when their paths cross? Answer in units of mi.
Click to expand...
If all else fails, you can just run through this one. :wink:

Start

-6.24 mi
+3.43mi

Run One Hour

-6.24 mi + 4.06 mi = - 2.81mi
+3.43 mi - 4.99 mi = -1.56 mi

It appears another hour will be too much.

You can set up equations to descibe the runners' locations.

-6.24 mi + t*(4.06 mi) = Position of Runner 1
+3.43 mi - t*(4.99 mi) = Position of Runner 2

...and ask when they will be the same.

-6.24 mi + t*(4.06 mi) = 3.43 mi - t*(4.99 mi)

Solve for 't' and you will know WHEN they run into each other. Use either formula to determine WHERE they ran into each other.
 
I don't understand your math on this one. What is the 4.06 mi and 4.99 miles? Also to solve for t at the bottom I cannot figure out what my time is? One hour?
 
I am guessing here because I still don't know what the 4.06 and 4.99 is?
So I did the problems like this because I don't really understand your reply.

-6.24 mi + 60m(4.06mi)= 237.36

3.43 mi+ 60M(4.99 mi)= 295.97

so in one hour this is where they will be? But is this telling me what the displacement of runner b from the flagpole when their paths cross?
 
Good idea to set up a similar but SIMPLER problem, like:

A@(6mph)>..............18 miles.............F.........10 miles.............<(@8mph)B

Solve that; the steps will then be same for your problem...
 
staceyrho said:
I have a problem, and I need to find the displacement of a runner. Here is the question:
Runner A is initially 6.24 mi W of a flagpole and is running with a constant velocigy of 4.06 mi/h dye east. Runner B is intitally 3.43 mi E of the flagpole abd us running with a constand velocity of 4.99 mi/h due W. Consider East to be the positive direction.
What is the displacement of runner B from the flagpole when their paths cross? Answer in units of mi.
Click to expand...

since A starts 6.24 miles west of the pole and B starts 3.43 miles east of the pole, then A and B are 6.24 + 3.43 = 9.67 miles apart when they start.

let x = the distance B travels until he/she meets A
9.67 - x = the distance A travels until she/he meets B

assuming they start at the same time, both runners will have run for the same amount of time when they meet ... let t = time both run in hrs

rate*time = distance

for runner B ...
(4.99 mph)(t hrs) = x miles

for runner A ...
(4.06 mph)(t hrs) = 9.67-x miles

divide the 1st equation by the 2nd ... (notice that the "t" gets eliminated)

(4.99 mph)(t hrs) = x miles
-------------------------------------
(4.06 mph)(t hrs) = 9.67-x miles

4.99/4.06 = x/(9.67 - x)

4.06x = 4.99(9.67 - x)

4.06x = 48.2533 - 4.99x

9.05x = 48.2533

x = 5.33 miles

runner B is 5.33 miles west of where he/she started from
 
You must log in or register to reply here.
Top