Question

The position of a cat running from a dog down a dark alley is given by the values of the table.


t(seconds)

0

1

2

3

4

5



s(feet)

0

7

29

61

88

104

A. Find the average velocity of the cat (ft/sec) for the time period beginning when t=2 and lasting
1)3s
2)2s

Answer 1

@tkhunny

Answer 2

b). Draw the graph of the function for yourself and estimate the instantaneous velocity of the cat (ft/sec) when t=2

Answer 3

\(Aveage\;Velocity = \dfrac{Total\;Displacement}{Total\;Time}\)

Answer 4

Yes, I got this formula but I am not sure how i apply for it..

Answer 5

I thought (104-29)/3=25.

Answer 6

but it seems that its wrong?

Answer 7

You have each position and each time. What are you missing?

Starting at t = 0
0 0
1 7 Avg = 7/1
2 29 Avg = 29/2
3 61 Avg = 61/3
4 88 Avg = 88/4
5 104 Avg = 104/5

Starting at t = 1
1 0
2 22 Avg = 22/1
3 54 Avg = 54/2
4 81 Avg = 81/3
5 97 Avg = 97/4

Starting at t = 2
2 0
3 32 Avg = 32/1
4 59 Avg = 59/2
5 75 Avg = 75/3

etc.

I did that awfully quickly, so you may need to check the arithmetic.

Answer 8

mm im still confused.

Answer 9

how did you get t=0?

Answer 10

Seriusly? You have it exactly correct. Perhaps you are struggling with the simplicity? Just go with it when you are done.

t = 0 is where you start.

Answer 11

so my euqation is right?
(104-29)/3

Answer 12

Well, that IS the last one I included in my length listing.

Did you start at t = 2? 29
Did you end 3 seconds later at t = 5? 104
How far is that? 104 - 29 = 75
How long did that take? 5 - 2 = 3

Really, you're done.

Answer 13

the fitst answer I got is 25. and its right. then i tried next one. it did not work

Answer 14

and also 5-2. How did you get that?

Answer 15

for 2) 2s. I did 104-29/2

Answer 16

Start at 2 and last 3 seconds.

1) Why would you EVER write that? You have forgotten your Order of Operations.

Start at 2 and last 2 seconds is (88-29)/2

Answer 17

thanks for taking your time

Answer 18

what about start at 2 and last 1 second ?

Answer 19

(61-29)/1

Are you just not understanding the chart? I've seen you do WAY trickier stuff. Something funny going on with this one.

Answer 20

I dont really understand velocity equation to this question. how do it do? B. Draw the graph of the function for yourself and estimate the instantaneous velocity of the cat (ft/sec) when t=2 I know that lim S(t)-S(a)\B-a/

Answer 21

mmm its confusing.

Answer 22

You were just forced to do a very large increment derivative experiment.

2==> 5 gives 25
2==> 4 gives 29.5
2==> 3 gives 32

What do you suppose might be the limit of hte process as the time period continues to shrink? This is EXACTLY the limiting process that produces the derivative.

Answer 23

tthe smaller number gets bigger.

Answer 24

Whoops!! Huge apologies. I just realized you have a very similar screen name as someone else. I was totally thinking you were someone else. Sorry about that. Let's try again.

Have you met the definition of the derivative? Usually somehting like this: \(\lim\limits_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h}\)

Answer 25

Its fine. aboustely, yes I know that equation. how do I use this for qustion b?

Answer 26

This is what we are doing, but with REALLY LARGE '\(h\)'.

Start at t = 2 and go for 3 seconds. This x = 2 and h = 3 \(\dfrac{104 - 29}{3} = 25\)
Start at t = 2 and go for 2 seconds. This x = 2 and h = 2 \(\dfrac{88 - 29}{2} = 29.5\)
Start at t = 2 and go for 1 second. This x = 2 and h = 1 \(\dfrac{61 - 29}{1} = 32\)

Are you seeing the pattern? The application of that Derivative-like structure?

Answer 27

Oh I got it now! so t=2 second and 3 to next is 104!

Answer 28

See how much better it gets when I'm talking to the right person?

Now, our task is to estimate the "Instantaneous Velocity" at t = 2. We've been using very large \(h\). It's time to start using a smaller value. Any ideas?

Answer 29

hahaha. mm near to 2? limit x=>2. therefore around 1.999?

Answer 30

Maybe. I have a weird idea.

x = 2 h = 3 gives 25
x = 2 h = 2 gives 29.5. That increased by 4.5 (29.5 - 25)
x = 2 h = 1 gives 32. That increased by 2.5 (32-29.5)
x = 2 h = 0 might give what? Or, in other words, it might increase by what from h = 1?

Answer 31

32?

Answer 32

Much smaller. Look at the values right after "increased by".

4.5
2.5
?? -- What might be a rational next value?

Answer 33

0.5?

Answer 34

Exactly what I had in mind.

This pattern should then look reasonable.

x = 2, h = 3 gives 25
x = 2, h = 2 gives 29.5. That increased by 4.5 (29.5 - 25)
x = 2, h = 1 gives 32. That increased by 2.5 (32-29.5)
x = 2, h = 0 might increase by 0.5 from x = 0, h = 1. That makes it 32.5.

I think we're done!

What we just did was apply the definition of a derivative in a creative way!

Answer 35

Draw the graph of the function for yourself and estimate the instantaneous velocity of the cat (ft/sec) when t=2
for this answer is..?

Answer 36

thats cool how its increased by 0.5

Answer 37

We just built a little model and used the model to estimate the value. A wonderful and useful idea.

You're going to have to do the graph estimation on your own. Draw the best pciture you can. Digital graphics would be beneficial. Try to trace a SMOOTH curve between all the points. Do your best to estimate the slope AT t = 2. It should be pretty steep!

There is another important estimate. It would be the average around the desired value. If we want to estimate it AT t = 2, use the average from 1 to 3.

x = 1, h = 2 gives (61-7)/2 = 54/2 = 27

Well, it is just a rough estimate. I certainly think the 32.5 is a better estimate.

Answer 38

Thank you soooooo much!!