Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

oh because i have seen this before!

Answer 8

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

Answer 9

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

Answer 10

\[y(4)-y(1)\] 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

Answer 11

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

Answer 12

ok I get the concepts thanks!