math456
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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so the distance covered / the time interval is what you're looking for...
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do you know how to find the distance covered between t=2 and t=2.5?
math456
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The average velocity!
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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
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yes
math456
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So in this case is .5 sec= 0?
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no.
math456
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And 0.05=7.2
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find y(2) find y(2.5)
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then find the difference between those two heights
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then divide by the time interval (.5s)
math456
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But 2.5 is not in the question..
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k
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2 + .5 = ?
math456
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2.5 ohh gotch ya so all I do is divide by the 0.5, 0.05?
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So for 0.5 it's 32?
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let me check
math456
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Okay
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yep
math456
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And for 0.05 is 320?
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don't think so... how'd you get that
math456
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Y(2)-y(2.5)/0.05
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naw...
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y(2) and y(2.05)
math456
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U r adding the values to 2.0?
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height at 2 seconds and height .05 seconds later ( y(2.05) )
math456
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so for 0.01 second its y(2)-y(2.01)
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\[(d _{f} - d _{i})/(t _{f} -t _{i})\]
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yep
math456
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okay how abt if i want to find the estimate instantaneous velocity when t=2
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sorry t=1*
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if you're allowed to use calculus you just take the derivative...
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if not, you take the limit of the avg. velocity expression as t->0
math456
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ummm not sure what u meant so when t=2 its t>0
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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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something like
\[V(t) = V _{o} +at\]
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seen that before?
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can i just do like 32-14.76M0.5-0.05N
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yea i hv seen it
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sorry here m is the divide sign
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The three ways I mentioned are the ways to do it, not sure what you typed there or what you're trying to do.
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You all good on this? Or do you have another question?
math456
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how how can i take the limit of avg. velocity at t>0?
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@Algebraic!
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like:
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for a specific value of t?
math456
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t=2
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I'll use the sketch pad to show you
math456
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okay, thanks
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|dw:1347077529182:dw|
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that's the first term...
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hmmm how do I find the delta t value?
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|dw:1347077587650:dw|
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that's the second term
math456
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okay.
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all that gets divided by delta t
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then you take the limit as delta t ->0
math456
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so the limit as x approaches to 0 is -16
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for inst. velocity?
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should be -24
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let me check again..
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okay got it nw..
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cool;)
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thanks for ur help..
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No problem!