If a ball is thrown straight up into the air with an initial velocity of 85ft/s its height in feet after t seconds is given by y=85t - 16t^2. Find the average velocity for the time period beginning when t=1 and lasting 0.1, 0.01, 0.001

Respuesta :

Answer:

[tex]51.4\ \text{m/s}[/tex]

[tex]52.84\ \text{m/s}[/tex]

[tex]52.984\ \text{m/s}[/tex]

Step-by-step explanation:

The function of displacement is

[tex]y=85t-16t^2[/tex]

[tex]\text{Average velocity}=\dfrac{\text{Total displacement}}{\text{Time elapsed}}[/tex]

Lasting 0.1 s

[tex]v_1=\dfrac{y(1.1)-y(1)}{1.1-1}\\\Rightarrow v_1=\dfrac{85\times 1.1-16\times 1.1^2-(85-16)}{0.1}\\\Rightarrow v_1=51.4\ \text{m/s}[/tex]

The average velocity is [tex]51.4\ \text{m/s}[/tex]

Lasting 0.01 s

[tex]v_2=\dfrac{y(1.01)-y(1)}{1.01-1}\\\Rightarrow v_2=\dfrac{85\times 1.01-16\times 1.01^2-(85-16)}{0.01}\\\Rightarrow v_2=52.84\ \text{m/s}[/tex]

The average velocity is [tex]52.84\ \text{m/s}[/tex]

Lasting  0.001 s

[tex]v_3=\dfrac{y(1.001)-y(1)}{1.001-1}\\\Rightarrow v_3=\dfrac{85\times 1.001-16\times 1.001^2-(85-16)}{0.001}\\\Rightarrow v_3=52.984\ \text{m/s}[/tex]

The average velocity is [tex]52.984\ \text{m/s}[/tex]