Question

Velocity questions

Answer 1

average velocity is just the slope between the endpoints.

Answer 2

So then box 2 and box 4 should both be 0.25?

Answer 3

dunno, i havent done the calculations ...

Answer 4

Well we just get the answers by plugging it into the equation?
s=t^2-7t+15

interval [3.5,4]
s(3.5) = (3.5)^2- 3.5(7)+15
and then do the same for 4 and the difference is 0.25...?

Answer 5

Except I tried putting in 0.25 and it didn't accept it.

Answer 6

Nvm. Just need hlep on the last one now.

Answer 7

Have you learned derivative shortcuts yet? :d

Answer 8

Or we still using the ole clunky limit definition for instantaneous change? :d

Answer 9

Just started derivatives o.o

Answer 10

\[\large\rm s(t)=t^2-7t+15\]
This s' will give us a function for instantaneous rate of change,\[\large\rm s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}\]As a final step, we'll plug t=4 into it.

Answer 11

Actually, plugging t=4 in right away might be a smarter move,
simplifies some of the algebra.

Answer 12

\[\large\rm s(4)=3\]

We'll also need this,\[\large\rm s(\color{orangered}{t+h})=(\color{orangered}{t+h})^2-7(\color{orangered}{t+h})+15\]which we will also evaluate at t=4,\[\large\rm s(\color{orangered}{4+h})=(\color{orangered}{4+h})^2-7(\color{orangered}{4+h})+15\]

Answer 13

\[\large\rm s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}\]
\[\large\rm s'(4)=\lim_{h\to0}\frac{s(4+h)-s(4)}{h}\]
\[\large\rm s'(4)=\lim_{h\to0}\frac{(4+h)^2-7(4+h)+15-3}{h}\]And then simplify, ya? :d
confusing? sup?

Answer 14

nah i'm following. Just lots of numbers xD

Answer 15

It all simplifies to 1, no? o.0

Answer 16

ooo that sounds right! :O
yay good job \c:/

Answer 17

yaaay! thank youu :)