If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t second is given by y=95t−16t2. Find the average velocity for the time period beginning when t=1 and lasting(i) 01 seconds:(ii) 001 seconds:(iii) 0001 seconds:Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1.

Respuesta :

Answer:

Instantaneous velocity is [tex]v=63[/tex] at t=1.

Explanation:

The height in feet after t second is given by y(t)=95t−16t2.

Average velocity is defined by:

[tex]v_{ave}=\frac{x_{f} - x_{i} }{t_{f} - t_{i}  }[/tex].

i)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.1\\x_{f}=y(1+0.1)=85.14\\[/tex]

⇒ [tex]v_{ave}=61.4[/tex].

ii)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.01\\x_{f}=y(1+0.01)=79.6284\\[/tex]

⇒ [tex]v_{ave}=62.84[/tex].

iii)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.001\\x_{f}=y(1+0.001)=79.062984\\[/tex]

⇒ [tex]v_{ave}=62.98[/tex].

Instantaneos velocity is defined by: [tex]v=\lim_{\triangle t \to 0} \frac{\triangle x}{\triangle t}[/tex]

So we have seen "manually" that when [tex]\triangle t \rightarrow 0[/tex], [tex]v \rightarrow 63[/tex].

So instantaneous velocity must be [tex]v=63[/tex] at t=1.