Help with this subbing equation to another equation question

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Runner A covers a distance of 60 meters while running at a constant speed of 6.0 m/s for 10 seconds. The discussion involves calculating the distance runner B travels as they accelerate from 4.0 m/s to 10.0 m/s over the same time period, which is determined to be 70 meters. There is confusion about the use of the 0.5 in the equations and whether it acts as a coefficient. The user contemplates whether to start a new thread for further questions or continue in the current discussion. The conversation also includes a personal note about the user's delay due to an unrelated task.
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Homework Statement



runner A runs at 6.0 m/s for 10s. Runnber B accelerates from 4.0 m/s to 10.0 m/s. steadily. in 10s.
how far does runner a go?

Homework Equations



v2=v1+at
d=v1*t+0.5a*t^2
d=Vavg*t

The Attempt at a Solution


ok this is how much I've worked it out so far
but I'm stuck now
v2=v1+at
a=(v2-v1)/2
v1=6m/s
t=10s
d=v1*t+0.5at^2
d=6*10+0.5((v2-6)/10)*10^2
does 0.5 multiply the denominator?
am i even doing this right?
 
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runner A runs at 6.0 m/s for 10s.
d = vt = 6*10 = 60 m
 
so... >_> it was that simple.. >< But does the 0.5 multiply the denominator? Also is the 0.5 in this case the coefficient? Is that a coefficient?
 
Hmm, too simple. Is there a part (b)? It might, for example, ask how far ahead one runner is after the 10 seconds.
 
Delphi51 said:
d = vt = 6*10 = 60 m

Delphi51 said:
Hmm, too simple. Is there a part (b)? It might, for example, ask how far ahead one runner is after the 10 seconds.

Hm well part b was How far does runner B go? Found out it was 70

Now I'm stuck with this question. Do i start a new thread or do i post it here and just edit it?
 
I agree with the 70.
Sorry about the delay; I had to lop two 6 inch branches off a tree 25 feet up!
Success, except I crashed one of them on my gate and damaged it.
 
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