Answer:
Our equation for the height is:
y(t) = 275 - 16*t^2.
a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:
![AV = \frac{y(t2) - y(t1)}{t2 - t1}](https://tex.z-dn.net/?f=AV%20%3D%20%5Cfrac%7By%28t2%29%20-%20y%28t1%29%7D%7Bt2%20-%20t1%7D)
Then:
i) t1 = 4s, t2 = 4s + 0.1s = 4.1s
The average velocity is:
![AV = \frac{(275 - 16*4.1^2) - (275 - 16*4^2)}{4.1 - 4} = \frac{16(4^2 - 4.1^2)}{0.1} = -129.6](https://tex.z-dn.net/?f=AV%20%3D%20%5Cfrac%7B%28275%20-%2016%2A4.1%5E2%29%20-%20%28275%20-%2016%2A4%5E2%29%7D%7B4.1%20-%204%7D%20%3D%20%5Cfrac%7B16%284%5E2%20-%204.1%5E2%29%7D%7B0.1%7D%20%3D%20-129.6)
And the units will be ft/s, so the average speed is:
-129.6 ft/s
The minus sign is because te pebble is falling down.
ii) t1 = 4s, t2 = 4s + 0.05s = 4.05s
The average velocity is:
![AV = \frac{(275 - 16*4.05^2) - (275 - 16*4^2)}{4.05 - 4} = \frac{16(4^2 - 4.05^2)}{0.05} = -128.8](https://tex.z-dn.net/?f=AV%20%3D%20%5Cfrac%7B%28275%20-%2016%2A4.05%5E2%29%20-%20%28275%20-%2016%2A4%5E2%29%7D%7B4.05%20-%204%7D%20%3D%20%5Cfrac%7B16%284%5E2%20-%204.05%5E2%29%7D%7B0.05%7D%20%3D%20-128.8)
So the average speed is -128.9 ft/s
iii) t1 = 4s, t2 = 4s + 0.01s = 4.01s
The average speed is:
![AV = \frac{(275 - 16*4.01^2) - (275 - 16*4^2)}{4.01 - 4} = \frac{16(4^2 - 4.01^2)}{0.01} = -128.16](https://tex.z-dn.net/?f=AV%20%3D%20%5Cfrac%7B%28275%20-%2016%2A4.01%5E2%29%20-%20%28275%20-%2016%2A4%5E2%29%7D%7B4.01%20-%204%7D%20%3D%20%5Cfrac%7B16%284%5E2%20-%204.01%5E2%29%7D%7B0.01%7D%20%3D%20-128.16)
The average speed is -128.16 ft/s.
b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.
v(t) = dy(t)/dt.
v(t) = -2*16*t + 0
Then the velocity at t = 4s is:
v(4s) = -32*4 = -128
The instantaneous velocity at t = 4s is -128 ft/s.