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:
-2.79 m/s²
Explanation:
Given:
v₀ = 20 m/s
v = 11 m/s
Δx = 50 m
Find: a
v² = v₀² + 2aΔx
(11 m/s)² = (20 m/s)² + 2a (50 m)
a = -2.79 m/s²
Round as needed.
Answer:
x = 5.79 m
Explanation:
given,
mass of the car = 39000 Kg
spring constant = 5.7 x 10⁵ N/m
acceleration due to gravity = 9.8 m/s²
height of the track = 25 m
length of spring compressed = ?
using conservation of energy
potential energy is converted into spring energy
x = 5.79 m
the spring is compressed to x = 5.79 m to stop the car.
Answer:
The answer is 11N to the right
Explanation:
Because 4N-3N= 1N
Therefore, 12N-1N=11N
The netforce is 11N to the right, because the greatest force is 12N to the right so it is more likely that the object is being pulled to the right.
Answer:
D.amplifying sound vibrations from the eardrum
this is correct
