Question 343639: Larry starts cycling at 20 mi/hr. One hour later, Rebecca starts cycling from the same place at 25 mi/hr, going in the same direction on the same road as Larry. How long will it take Rebecca to catch up to Larry?
*Someone told me to do
20T = 25(T-1)
20T = 25T - 25
20T - 25T = -25
-5T = -25
T = -25/-5
T = 5 hours.
I am not sure if this is right? If it is right, I don't understand what everything stands for or the formula used. I need an explanation please.
Found 2 solutions by ankor@dixie-net.com, scott8148:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Larry starts cycling at 20 mi/hr. One hour later, Rebecca starts cycling from the same place at 25 mi/hr, going in the same direction on the same road as Larry. How long will it take Rebecca to catch up to Larry?
:
This is right, this equation makes use of the fact that when R catch L then
they will have traveled the same distance.
Dist = speed * time
:
T = L's travel time. (dist = 20T)
R started 1 hour later so
(T-1) = R's travel time. dist = (25(t-1))
:
With this in mind, the equation is:
L's dist = R's dist
20T = 25(T-1)
20T = 25T - 25
20T - 25T = -25
-5T = -25
T = -25/-5
T = 5 hours.
YOu solved for T correctly, you can prove this by finding the dist each has
traveled, they should be equal
L's dist: 20*5 = 100 mi
R's dist: 25(5-1) = 100 mi also
:
:
Did this shed some light on this kind of problem, any questions? use the comment feature.
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! Rebecca catches Larry when they have both traveled the same distance
distance equals rate multiplied by time
Rebecca's rate is 25 mph and her time is one hr less than Larry's
Larry's rate is 20 mph
the T in the equations represents Larry's time
depending on your viewpoint, you might say that Rebecca's time to catch Larry doesn't start until Rebecca starts
___ so the answer is either 4 hr or 5 hr
hope this helps
|
|
|
