Answer:
The acceleration of the object is 9.3 m/s²
Explanation:
For a straight movement with constant acceleration, this equation for the position applies:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:
x₁ = x0 + v0 t + 1/2 a (1 s)²
x₂ = x0 + v0 t + 1/2 a (2 s)²
x₂ - x₁ = 14 m
we know that the object starts from rest, so v0 = 0
substracting both equations of position we will get:
x₂ - x₁ = 14
x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m
x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m
2 a s² - 1/2 a s² = 14 m
3/2 a s² = 14 m
a = 14 m / (3/2 s²) = <u>9.3 m/s² </u>
D reference point. i hope that helped
B. because it's kind of like a fulcrum.
Answer:
11.7 s
Explanation:
In this problem, the rocket is moving in a uniform accelerated motion. We have the following data:
d = 223 m, the distance that the sled has to cover
, the acceleration of the rocket
We can use therefore the following SUVAT equation:
where
d is the distance
u = 0 is the initial velocity of the sled (it starts from rest)
t is the time
a is the acceleration
Re-arranging the equation and substituting the numbers, we find the time it takes for the rocket to cross the field: