Question

Calculus help I'll post the pic

Answer 1

5

Answer 2

Given a position function \(y(t)\), the average velocity over some time interval \([a,b]\) is
\[\frac{y(b)-y(a)}{b-a}\]

Answer 3

And to find the instantaneous velocity, it looks like you should use the results for part a, i-iv, to determine a pattern.

Answer 4

I use the formula you gave me?

Answer 5

I just didnt know the average part like sith did :P But this problem is now in goodf hands xD

Answer 6

@angelita321, for example, the average velocity for the first interval, \([2,2.5]\), is
\[\frac{y(2.5)-y(2)}{2.5-2}=\frac{(40(2.5)-16(2.5)^2)-(40(2)-16(2)^2)}{0.5}=\cdots\]

Answer 7

So for I) its like [2,.5]?

Answer 8

No each interval begins with \(t=2\) and ends with \(t=2+\cdots\), where \(\cdots \) is the time given in each respective question.

Answer 9

So how would I do I?

Answer 10

(i) The average velocity for the time period beginning when \(t=2\) and lasting \(0.5\) second is
\[\frac{y(2.5)-y(2)}{2.5-2}=\frac{(40(2.5)-16(2.5)^2)-(40(2)-16(2)^2)}{0.5}=\cdots\]
(ii) ... and lasting 0.1 second is
\[\frac{y(2.1)-y(2)}{2.1-2}=\frac{(40(2.1)-16(2.1)^2)-(40(2)-16(2)^2)}{0.1}=\cdots\]
(iii) ...

See where this is heading? I'm leaving the rest to you.

Answer 11

Thanks!! I got that. Do I solve it like this for #6?

Answer 12

Yes, for 6 you're explicitly given the interval, but it's the same process.

Answer 13

Okkay. Thanks !!!! (:

Answer 14

You're welcome!

Answer 15

Have you noticed this type of question is not usually in the physics text book? :P

Answer 16

Thats true

Answer 17

But this is a good training, makes you analyze physics more mathematically.