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Mathematics 21 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

The average velocity!

OpenStudy (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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

So in this case is .5 sec= 0?

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

And 0.05=7.2

OpenStudy (anonymous):

find y(2) find y(2.5)

OpenStudy (anonymous):

then find the difference between those two heights

OpenStudy (anonymous):

then divide by the time interval (.5s)

OpenStudy (anonymous):

But 2.5 is not in the question..

OpenStudy (anonymous):

k

OpenStudy (anonymous):

2 + .5 = ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

So for 0.5 it's 32?

OpenStudy (anonymous):

let me check

OpenStudy (anonymous):

Okay

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

And for 0.05 is 320?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Y(2)-y(2.5)/0.05

OpenStudy (anonymous):

naw...

OpenStudy (anonymous):

y(2) and y(2.05)

OpenStudy (anonymous):

U r adding the values to 2.0?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sorry t=1*

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

seen that before?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yea i hv seen it

OpenStudy (anonymous):

sorry here m is the divide sign

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@Algebraic!

OpenStudy (anonymous):

like:

OpenStudy (anonymous):

for a specific value of t?

OpenStudy (anonymous):

t=2

OpenStudy (anonymous):

I'll use the sketch pad to show you

OpenStudy (anonymous):

okay, thanks

OpenStudy (anonymous):

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