Respuesta :
Answer:
a) Average velocity at 0.1 s is 696 ft/s.
b) Average velocity at 0.01 s is 7536 ft/s.
c) Average velocity at 0.001 s is 75936 ft/s.
Step-by-step explanation:
Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by [tex]y = 70t-16t^2[/tex].
To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.
Solution :
a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.
[tex](\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}[/tex]
[tex](\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}[/tex]
[tex](\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}[/tex]
[tex](\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}[/tex]
[tex](\text{Average velocity})_{0.1\ s}=696\ ft/s[/tex]
b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.
[tex](\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}[/tex]
[tex](\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}[/tex]
[tex](\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}[/tex]
[tex](\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}[/tex]
[tex](\text{Average velocity})_{0.01\ s}=7536\ ft/s[/tex]
c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.
[tex](\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}[/tex]
[tex](\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}[/tex]
[tex](\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}[/tex]
[tex](\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}[/tex]
[tex](\text{Average velocity})_{0.001\ s}=75936\ ft/s[/tex]
