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?
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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
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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
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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?
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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?
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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?
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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
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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...
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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
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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:
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