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anonymous
 3 years ago
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= 40t16t^2. Find the avg velocity for the time period beginning when t=2 & lasting 0.5 sec, 0.05 sec?
anonymous
 3 years ago
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= 40t16t^2. Find the avg velocity for the time period beginning when t=2 & lasting 0.5 sec, 0.05 sec?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the distance covered / the time interval is what you're looking for...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you know how to find the distance covered between t=2 and t=2.5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The average velocity!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yea it's like u plug in the 2 in the equation and then distance covered in 2distance coverd in 2.5/22.5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So in this case is .5 sec= 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0find y(2) find y(2.5)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then find the difference between those two heights

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then divide by the time interval (.5s)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But 2.5 is not in the question..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02.5 ohh gotch ya so all I do is divide by the 0.5, 0.05?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0don't think so... how'd you get that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0U r adding the values to 2.0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0height at 2 seconds and height .05 seconds later ( y(2.05) )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so for 0.01 second its y(2)y(2.01)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(d _{f}  d _{i})/(t _{f} t _{i})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay how abt if i want to find the estimate instantaneous velocity when t=2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you're allowed to use calculus you just take the derivative...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if not, you take the limit of the avg. velocity expression as t>0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ummm not sure what u meant so when t=2 its t>0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the third alternative is using the equation that you might've been given where the derivative is already taken for you...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0something like \[V(t) = V _{o} +at\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can i just do like 3214.76M0.50.05N

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry here m is the divide sign

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The three ways I mentioned are the ways to do it, not sure what you typed there or what you're trying to do.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You all good on this? Or do you have another question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how how can i take the limit of avg. velocity at t>0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for a specific value of t?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll use the sketch pad to show you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347077529182:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the first term...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm how do I find the delta t value?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347077587650:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the second term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0all that gets divided by delta t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then you take the limit as delta t >0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the limit as x approaches to 0 is 16
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