I think the answer you're looking for is -are randomly oriented. if not sorry... i tried.
V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
Answer:
Δd = 
Explanation:
As 
, when the car is making full stop, 
. 
. Therefore,
Apply the same formula above, with 
and 
, and the car is starting from 0 speed, we have
As 
. After 
, the car would have traveled a distance of
Hence 
As 
we can simplify 
After t time, the train would have traveled a distance of 
Therefore, Δd would be 
I would use miles
Because miles will give you less of an answer
Answer:
Explanation: nigerian what
