Respuesta :
Answer:
a. 38.19m/s
b. 38.605m/s
c. 38.937m/s
d. 39.0117m/s
e. 39.01917m/s
Step-by-step explanation:
The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:
[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]
Where:
[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]
a. Let's find h(3) and h(4) using the data provided by the problem:
[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]
The average velocity over the interval [3, 4] is :
[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]
b. Let's find h(3.5) using the data provided by the problem:
[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]
The average velocity over the interval [3, 3.5] is :
[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]
c. Let's find h(3.1) using the data provided by the problem:
[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]
The average velocity over the interval [3, 3.1] is :
[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]
d. Let's find h(3.01) using the data provided by the problem:
[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]
The average velocity over the interval [3, 3.01] is :
[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]
e. Let's find h(3.001) using the data provided by the problem:
[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]
[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]
The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.
Given :
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].
a) [3,4]
At time t = 3 and 4, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]
b) [3,3.5]
At time t = 3 and 3.5, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]
c) [3,3.1]
At time t = 3 and 3.1, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]
d) [3,3.01]
At time t = 3 and 3.01, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]
e) [3,3.001]
At time t = 3 and 3.001, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]
For more information, refer to the link given below:
https://brainly.com/question/18153640
