When a ball is thrown vertically upward on the moon with a velocity of 25 ft/sec its height, y(t), in feet after t seconds is given by y(t) = 25t − 3t2 . Find the average velocity of the ball over the interval from 1 to 1 + h seconds, h = 0.

can't figure this question out, if somewhere could walk me through it I would appreciate it.

you are asked for \[\frac{y(1+h)-y(1)}{h}\]

when t = 1, y = 22 when t = 1+h, \[y=25(1+h)-3(1+h)^2=-3 h^2+19 h+22\]

that was the only hard part, the algebra. now \[\frac{y(1+h)-y(1)}{h}=\frac{-3h^2-19h+22-22}{h}=\frac{-3h^2+19h}{h}\]

cancel the h carefully, get \[-3h+19\] now if you want to replace h by 0 to get the instantaneous velocity (not the average) at t = 1 you get 19 feet per second

how did you automatically know how to set up that first equation?

oh because i have seen this before!

ok lets look carefully. it says Find the average velocity of the ball over the interval from 1 to 1 + h

suppose h was 3. then you would find the average over the interval from 1 to 4 to find it you use the slope \[\frac{y(4)-y(1)}{4-1}\]

\[y(4)-y(1)\] gives the distance traveled from 1 to 4 seconds. and 4 - 1 = 3 says it took 3 seconds to go that far, so average velocity would be distance traveled / time elapsed

strictly speaking what i am saying is false, because this thing goes up and then down, but the idea is right. it is like saying if at 1 pm you are at mile marker 40 on the highway and at 4 pm you are at mile marker 200 then you went 200 - 40 = 160 miles in 4 - 1 = 3 hours so your average speed is 160/3

ok I get the concepts thanks!

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