Question

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 h(t)=44t−0.83t2h(t)=44t-0.83t2. Find the average velocity over the given time intervals.
a. [3, 4]:
b. [3, 3.5]:
c. [3, 3.1]:
d. [3, 3.01]:
e. [3, 3.001]

Answer 1

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:

$v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}$

Where:

$x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval$

a. Let's find h(3) and h(4) using the data provided by the problem:

$h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f$

The average velocity over the interval [3, 4] is :

$v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s$

b. Let's find h(3.5) using the data provided by the problem:

$h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f$

The average velocity over the interval [3, 3.5] is :

$v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s$

c. Let's find h(3.1) using the data provided by the problem:

$h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f$

The average velocity over the interval [3, 3.1] is :

$v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s$

d. Let's find h(3.01) using the data provided by the problem:

$h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f$

The average velocity over the interval [3, 3.01] is :

$v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s$

e. Let's find h(3.001) using the data provided by the problem:

$h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f$

$v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s$