SOLUTION: suzy can run 4m/sec and tim can run 6m/sec. How far ahead of tim must suzy be to not to fall behind tim in the first 10 secondsof running? Use a graph to check ou answer?

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Question 150990: suzy can run 4m/sec and tim can run 6m/sec. How far ahead of tim must suzy be to not to fall behind tim in the first 10 secondsof running? Use a graph to check ou answer?
Found 2 solutions by ptaylor, ankor@dixie-net.com:
Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!

Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let x=distance ahead of Tim that Suzy must be to not to fall behind tim in the first 10 secondsof running
Distance Tim runs in the first 10 seconds=(6m/s)*10 sec=60 meters
Distance Suzy runs in the first 10 seconds plus her head start of x meters equals 4m/s*10 sec + x=40 meters plus x
Now we know that when the above two distances are equal, Tim will have caught up and in the next instance, Suzy will have fallen behind, so our equation to solve is:
60=40+x subtract 40 meters from each side
60-40=x
x=20 meters--------------------distance ahead that Suzy must be so as not to fall behind in the first 10 seconds
MAYBE YOU CAN DO THE GRAPH
CK
20+4*10=60
60=60
Hope this helps---ptaylor

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
suzy can run 4m/sec and tim can run 6m/sec. How far ahead of tim must suzy be to not to fall behind tim in the first 10 seconds of running? Use a graph to check your answer?
:
Let d = distance that T is ahead of S
:
Write a distance equation: Dist = speed * 10 sec
:
S's dist + head-start dist = T's dist
4(10) + d = 6(10)
40 + d = 60
d = 60 - 40
d = 20 m head start
:
"Use a graph to check your answer"
let x = time running
let y = distance ran
;
Suzy: y = 4x + 20; (green)
:
Tim: y = 6x; (purple)
:
y axis = dist, x axis = time

You can see that T (purple) passes S (green) at 10 sec
And when x=0, S has a 20m headstart