Respuesta :
With the average acceleration in each interval we can find that maximum acceleration occurs in the first interval and has a value of a = 16 m / s²
Average acceleration is defined as the change in velocity of a body in a given time interval
a = [tex]\frac{\Delta v}{\Delta t}[/tex]
where a is the acceleration, Δv and Δt the speed and time variations in the interval.
In the attachment we have a graph of the speed as a function of time where it is observed that the speed is reaching a maximum as time advances,
let's calculate the average acceleration for each interval
1) time between t₁ = 0 and t_f = 50 10⁻³ s for velocities v₁ = 0 and v_f = 0.8 m/s
a₁ = [tex]\frac{0.8 -0}{(50 - 0) 10^{-3}}[/tex]
a₁ = 16 m / s²
2) time between t₁ = 50 10⁻³and t_f = 70 10⁻³ s for velocities v₁ = 1.0 and
v_f = 0.8 m/s
a₂ = [tex]\frac{1.0 - 0.8}{(70-50) \ 10^{-3}}[/tex]
a₂ = 10 m / s²
3) time between t₁ = 70 10⁻³ s and t_f = 100 10⁻³ s for velocities v₁ = 0 and
v_f = 0.8 m/s
a₃ = [tex]\frac{1.3 - 1.0}{(100 -70) \ 10^{-3}}[/tex]
a₃ = 10 m / s²
4) time t₁ = 100 10⁻³ s and t_f = 125 10⁻³ s for speeds v₁ = 1.3 m / s and
v_f = 1.5 m / s
a₄ = [tex]\frac{ 1.5 - 1.3}{(125-100)10^{-3}}[/tex]
a₄ = 8 m / s²
5) time t₁ = 125 10⁻³ s and t_f = 150 10⁻³ s for speeds v₁ = 1.5 m / s and
v_f = 1.65 m / s
a₅ = [tex]\frac{1.65 - 1.5}{(150-125)10^{-3}}[/tex]
a₅ = 6 m / s²
6) time t₁ = 150 10⁻³ s and t_f = 200 10⁻³ ms for speeds v₁ = 1.65 m / s and
v_f = 1.8 m / s
a₆ = [tex]\frac{1.8 - 1.65}{(200 -150)10^{-3}}[/tex]
a₆ = 3 m / s²
7) time t₁ = 200 10⁻³ s and t-f = 230 10⁻³ s for speeds v₁ = 1.8 m / s and
v_f = 1.9 m / s
a₇ = [tex]\frac{1.9 - 1.8}{(230-200)10^{-3}}[/tex]
a₇ = 3 m / s
In conclusion, using the concepts of average acceleration we can find that maximum acceleration occurs in the first interval and has a value of
a = 16 m / s²
learn more about average acceleration here:
https://brainly.com/question/10064420
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