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: your correct answer is a i took the test
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Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Number 1. The medium around the wire
Answer:
The magnitude of the acceleration of the elevator is 0.422 m/s²
Explanation:
Lets explain how to solve the problem
Due to Newton's Law ∑ Forces in direction of motion is equal to mass
multiplied by the acceleration
We have here two forces 460 N in direction of motion and the weight
of the person in opposite direction of motion
The weight of the person is his mass multiplied by the acceleration of
gravity
→ W = mg , where m is the mass and g is the acceleration of gravity
→ m = 45 kg and g = 9.8 m/s²
Substitute these values in the rule above
→ W = 45 × 9.8 = 441 N
The scale reads 460 N
→ F = 460 N , W = 441 N , m = 45 kg
→ F - W = ma
→ 460 - 441 = 45 a
→ 19 = 45 a
Divide both sides by 45
→ a = 0.422 m/s²
<em>The magnitude of the acceleration of the elevator is 0.422 m/s²</em>
