K you have to be careful that you understand where your formulas are coming from so you know what they mean and when they apply. I usually just stick to coming up with the equations myself so I know what they mean. It's also less to remember.
The governing idea here is constant acceleration, so starting from that:
a(t) = a, where a is the constant acceleration value
v(t) = integral(a(t)) = a*t + v0 where v0 is the initial velocity
x(t) = integral(v(t)) = 1/2*a*t^2 + v0*t + x0 where x0 is the initial position
To get your formula, I set v0=x0=0 and call x(t)=d, I get:
d=1/2*a*t^2
which solves to:
t=(2*d/a)^(1/2)
Unfortunately though, you don't get to set v0=0, because they're clearly not starting the last fraction of the race at a dead stop.
Took me a while, but I got this one too.
The time after the 1/9 of the race is seven seconds after the time before. So you subtract seven from the "after" half of the equation, and add 8/9 as the distance for the "before" part. Or:
(2*d/a)^(1/2)-7=((2*8*d/9)/a)^(1/2)
Work with that to get the 7 alone, and you get:
((8/9)^(1/2))((2x/a)^(1/2))-((2x/a)^(1/2))=7
Pull out the ((2x/a)^(1/2)), and you get ((2x/a)^(1/2))(1-((8/9)^(1/2))=7
((2x/a)^(1/2)), if you remember from the beginning, is equal to time.
T*(1-sqr(8/9))=7
T=7/(sqr(8/9))
Work.
(this is online cause I'm workin' with a friend who wanted me to link it to them)
The next one stumping me is a stupid reimann sum. I know how to do reimann sums, and I've entered in the "correct" thing, but the online is saying my answer is wrong. It's converting from velocity to time, which might be my issue.
