anonymous
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
so the distance covered / the time interval is what you're looking for...
anonymous
  • anonymous
do you know how to find the distance covered between t=2 and t=2.5?
anonymous
  • anonymous
The average velocity!

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More answers

anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
So in this case is .5 sec= 0?
anonymous
  • anonymous
no.
anonymous
  • anonymous
And 0.05=7.2
anonymous
  • anonymous
find y(2) find y(2.5)
anonymous
  • anonymous
then find the difference between those two heights
anonymous
  • anonymous
then divide by the time interval (.5s)
anonymous
  • anonymous
But 2.5 is not in the question..
anonymous
  • anonymous
k
anonymous
  • anonymous
2 + .5 = ?
anonymous
  • anonymous
2.5 ohh gotch ya so all I do is divide by the 0.5, 0.05?
anonymous
  • anonymous
So for 0.5 it's 32?
anonymous
  • anonymous
let me check
anonymous
  • anonymous
Okay
anonymous
  • anonymous
yep
anonymous
  • anonymous
And for 0.05 is 320?
anonymous
  • anonymous
don't think so... how'd you get that
anonymous
  • anonymous
Y(2)-y(2.5)/0.05
anonymous
  • anonymous
naw...
anonymous
  • anonymous
y(2) and y(2.05)
anonymous
  • anonymous
U r adding the values to 2.0?
anonymous
  • anonymous
height at 2 seconds and height .05 seconds later ( y(2.05) )
anonymous
  • anonymous
so for 0.01 second its y(2)-y(2.01)
anonymous
  • anonymous
\[(d _{f} - d _{i})/(t _{f} -t _{i})\]
anonymous
  • anonymous
yep
anonymous
  • anonymous
okay how abt if i want to find the estimate instantaneous velocity when t=2
anonymous
  • anonymous
sorry t=1*
anonymous
  • anonymous
if you're allowed to use calculus you just take the derivative...
anonymous
  • anonymous
if not, you take the limit of the avg. velocity expression as t->0
anonymous
  • anonymous
ummm not sure what u meant so when t=2 its t>0
anonymous
  • anonymous
the third alternative is using the equation that you might've been given where the derivative is already taken for you...
anonymous
  • anonymous
something like \[V(t) = V _{o} +at\]
anonymous
  • anonymous
seen that before?
anonymous
  • anonymous
can i just do like 32-14.76M0.5-0.05N
anonymous
  • anonymous
yea i hv seen it
anonymous
  • anonymous
sorry here m is the divide sign
anonymous
  • 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
  • anonymous
You all good on this? Or do you have another question?
anonymous
  • anonymous
how how can i take the limit of avg. velocity at t>0?
anonymous
  • anonymous
@Algebraic!
anonymous
  • anonymous
like:
anonymous
  • anonymous
for a specific value of t?
anonymous
  • anonymous
t=2
anonymous
  • anonymous
I'll use the sketch pad to show you
anonymous
  • anonymous
okay, thanks
anonymous
  • anonymous
|dw:1347077529182:dw|
anonymous
  • anonymous
that's the first term...
anonymous
  • anonymous
hmmm how do I find the delta t value?
anonymous
  • anonymous
|dw:1347077587650:dw|
anonymous
  • anonymous
that's the second term
anonymous
  • anonymous
okay.
anonymous
  • anonymous
all that gets divided by delta t
anonymous
  • anonymous
then you take the limit as delta t ->0
anonymous
  • anonymous
so the limit as x approaches to 0 is -16
anonymous
  • anonymous
for inst. velocity?
anonymous
  • anonymous
should be -24
anonymous
  • anonymous
let me check again..
anonymous
  • anonymous
okay got it nw..
anonymous
  • anonymous
cool;)
anonymous
  • anonymous
thanks for ur help..
anonymous
  • anonymous
No problem!

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