Actually the position function with respect to time under constant acceleration is:
a=g
v=⌠g dt
v=gt+vi
s=⌠v
s=gt^2/2+vit+si
So if vi and si are zero then you just have:
s=gt^2/2
Notice that it is not gt^2 but (g/2) t^2
So the first term in any quadratic is half of the acceleration times time squared because of how the integration works out...
Anyway....
sf=(a/2)t^2+vit+si
(sf-si)-vit=a(t^2)/2
2(sf-si)-2vit=at^2
a=(2(sf-si)-2vit)/t^2 and if si and vi equal zero
a=(2s)/t^2
The answer is B, there’s no solution.
Format: y-y1 = m(x-x1)
Solution: B. y + 1.4 = 1.9(x - 4.5)
Answer:
13w-7
Step-by-step explanation:
-6w + (-8) + 1 + (-7w)
(-6w+-7w) + (-8 + 1)
-13w + (-7) or -13w - 7
