If a ball is thrown into the air with a v of 40 ft/s, it's height in ft t seconds late is given by y= 40t-16t^2. Find the avg velocity for the time period beginning when t=2 & lasting 0.5 sec, 0.05 sec?

- anonymous

- katieb

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- anonymous

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

- anonymous

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

- anonymous

The average velocity!

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## More answers

- anonymous

Yea it's like u plug in the 2 in the equation and then distance covered in 2-distance coverd in 2.5/2-2.5

- anonymous

yes

- anonymous

So in this case is .5 sec= 0?

- anonymous

no.

- anonymous

And 0.05=7.2

- anonymous

find y(2) find y(2.5)

- anonymous

then find the difference between those two heights

- anonymous

then divide by the time interval (.5s)

- anonymous

But 2.5 is not in the question..

- anonymous

k

- anonymous

2 + .5 = ?

- anonymous

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

- anonymous

So for 0.5 it's 32?

- anonymous

let me check

- anonymous

Okay

- anonymous

yep

- anonymous

And for 0.05 is 320?

- anonymous

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

- anonymous

Y(2)-y(2.5)/0.05

- anonymous

naw...

- anonymous

y(2) and y(2.05)

- anonymous

U r adding the values to 2.0?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yep

- anonymous

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

- anonymous

sorry t=1*

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

the third alternative is using the equation that you might've been given where the derivative is already taken for you...

- anonymous

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

- anonymous

seen that before?

- anonymous

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

- anonymous

yea i hv seen it

- anonymous

sorry here m is the divide sign

- anonymous

The three ways I mentioned are the ways to do it, not sure what you typed there or what you're trying to do.

- anonymous

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

- anonymous

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

- anonymous

- anonymous

like:

- anonymous

for a specific value of t?

- anonymous

t=2

- anonymous

I'll use the sketch pad to show you

- anonymous

okay, thanks

- anonymous

|dw:1347077529182:dw|

- anonymous

that's the first term...

- anonymous

hmmm how do I find the delta t value?

- anonymous

|dw:1347077587650:dw|

- anonymous

that's the second term

- anonymous

okay.

- anonymous

all that gets divided by delta t

- anonymous

then you take the limit as delta t ->0

- anonymous

so the limit as x approaches to 0 is -16

- anonymous

for inst. velocity?

- anonymous

should be -24

- anonymous

let me check again..

- anonymous

okay got it nw..

- anonymous

cool;)

- anonymous

thanks for ur help..

- anonymous

No problem!

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