Respuesta :
Answer:
Magnitude of Paula’s acceleration[tex]$a=39 \cdot 8m/sec^{2}$.[/tex]
Explanation:
• An object is said to be accelerated if there is a change in its velocity. At any point on a trajectory, the magnitude of the acceleration is given by the rate of change of velocity in both magnitude and direction at that point.
• To find the magnitude of her acceleration, use the formula: [tex]${v^2} = {u^2} + 2as$[/tex]
Where, is final velocity, is initial velocity, is acceleration and is displacement.
• Placing the value of the given initial velocity, [tex]$u=0m/s$[/tex], displacement, [tex]$s = 0 \cdot 201m$[/tex] and the final velocity,[tex]$v = 4m/s$[/tex] in the above formula.
[tex]\[\begin{align}& \therefore{v^2} = {u^2} + 2as \\& \Rightarrow {\left( 4 \right)^2} = 0+ 2\left( a \right) \left( { 0 \cdot 201} \right) \\& \Rightarrow 16 = 0 \cdot 402a \\& \Rightarrow a = 39 \cdot 8m/sec^{2} \\\end{align}\][/tex]
[tex]\[& \therefore{v^2} = {u^2} + 2as \\& \Rightarrow {\left( 4 \right)^2} = 0+ 2\left( a \right) \left( { 0 \cdot 201} \right) \\& \Rightarrow 16 = 0 \cdot 402a \\& \Rightarrow a = 39 \cdot 8m/sec^{2} \\\][/tex]
• Hence, magnitude of her acceleration, [tex]$a=39 \cdot 8m/sec^2$[/tex]
Learn more about acceleration here:
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The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/[tex]s^{2}[/tex].
Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.
To determine Paola's acceleration, the third equation of free fall is appropriate.
i.e [tex]V^{2}[/tex] = [tex]U^{2}[/tex] ± 2as
where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.
From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.
So that since Poala flies against gravity, then we have:
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] - 2as
0 = [tex](4)^{2}[/tex] - 2(a x 0.201)
= 16 - 0.402a
0.402a = 16
a = [tex]\frac{16}{0.402}[/tex]
= 39.801
a = 39.80 m/[tex]s^{2}[/tex]
Therefore Paola's acceleration is 39.80 m/[tex]s^{2}[/tex].
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