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) = 9.3 m/s²
The constant acceleration of the object is 9.33 m/s².
The given parameters;
The acceleration of the object is calculated by applying the second kinematic equation as follows;
s = ut + ¹/₂at²
where;
at t = 1.0 s
x₁ = 0 + ¹/₂(1²)a
x₁ = ¹/₂(a)
at t = 2.0 s
x₂ = 0 + ¹/₂(2²)a
x₂ = 2a
The change in position;
Δx = x₂ - x₁ = 14
14 = 2a - ¹/₂(a)
[tex]14 = \frac{3a}{2} \\\\3a = 2(14)\\\\3a = 28\\\\a = \frac{28}{3} = 9.33 \ m/s^2[/tex]
Thus, the constant acceleration of the object is 9.33 m/s².
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