Scientists have studied two species of sand lizards, the Mojave fringe-toed lizard and the western zebra-tailed lizard, to understand the extent to which the different structure of the two species' toes is related to their preferred habitats-fine sand for the Mojave lizard and coarse sand for the zebra-tailed lizard. (Figure 1) shows a somewhat simplified velocity-versus-time graph for the Mojave fringe-toed lizard. Figure1 of 1 The plot shows velocity v subscript x as a function of time t. Time is measured from 0 to 250 milliseconds on the x-axis. Velocity is measured from 0 to 2.0 meters per second on the y-axis. Velocity equals 0 at zero time. Then it increases linearly to 0.8 meter per second at 50 milliseconds. Then it increases at a decreasing rate forming the curve concave down that passes 1.0 meter per second at 70 milliseconds, 1.3 meters per second at 100 milliseconds, 1.5 meters per second at 125 milliseconds, 1.65 meters per second at 150 milliseconds, 1.8 meters per second at 200 milliseconds, and 1.9 meters per second at 230 milliseconds. Part A Estimate the maximum acceleration of the lizard in both m/s2 and g's.

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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