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so the distance covered / the time interval is what you're looking for...

do you know how to find the distance covered between t=2 and t=2.5?

The average velocity!

yes

So in this case is .5 sec= 0?

no.

And 0.05=7.2

find y(2) find y(2.5)

then find the difference between those two heights

then divide by the time interval (.5s)

But 2.5 is not in the question..

2 + .5 = ?

2.5 ohh gotch ya so all I do is divide by the 0.5, 0.05?

So for 0.5 it's 32?

let me check

Okay

yep

And for 0.05 is 320?

don't think so... how'd you get that

Y(2)-y(2.5)/0.05

naw...

y(2) and y(2.05)

U r adding the values to 2.0?

height at 2 seconds and height .05 seconds later ( y(2.05) )

so for 0.01 second its y(2)-y(2.01)

\[(d _{f} - d _{i})/(t _{f} -t _{i})\]

yep

okay how abt if i want to find the estimate instantaneous velocity when t=2

sorry t=1*

if you're allowed to use calculus you just take the derivative...

if not, you take the limit of the avg. velocity expression as t->0

ummm not sure what u meant so when t=2 its t>0

something like
\[V(t) = V _{o} +at\]

seen that before?

can i just do like 32-14.76M0.5-0.05N

yea i hv seen it

sorry here m is the divide sign

You all good on this? Or do you have another question?

how how can i take the limit of avg. velocity at t>0?

like:

for a specific value of t?

t=2

I'll use the sketch pad to show you

okay, thanks

|dw:1347077529182:dw|

that's the first term...

hmmm how do I find the delta t value?

|dw:1347077587650:dw|

that's the second term

okay.

all that gets divided by delta t

then you take the limit as delta t ->0

so the limit as x approaches to 0 is -16

for inst. velocity?

should be -24

let me check again..

okay got it nw..

cool;)

thanks for ur help..

No problem!