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avatar+209 

s=s(t)=sqrt(4t)     Find the average velocity of the object on [1,4]

 

Please help, and thank you to those who do.

 Feb 2, 2016

Best Answer 

 #3
avatar+128089 
+15

I'm confused, Yura_chan.....if   s(t)  =  sqrt(4t)   is the position function.....then....the average velocity  =  

 

[change in position]  / time traveled 

 

The  position  at t = 4 = sqrt(4*4)  = 4 cm

 

And the position at t = 1 = sqrt(4*1)  =  2cm

 

So ......

 

It moves 2cm  in 3 seconds    =  2/3  cm  per second....

 

 

cool cool cool

 Feb 2, 2016
 #1
avatar+128089 
+10

Avg velocity  = [ s(b)  - s(a)] / [ b  - a  ]      where b = 4 and a = 1    .....so we have...

 

[ sqrt(4*4)  -  sqrt(4*1)]  /  [ 4 - 1  ]  =

 

[ sqrt(16)  -  sqrt(4)]  /  [ 4 - 1  ]  =

 

[4  -  2 ] / 3  =

 

2  / 3    

 

 

 

cool cool cool

 Feb 2, 2016
 #2
avatar+209 
+10

I thought I'd have to use the formula f(x)-f(c)/x-c 
Is what I've been doing correct?

I think that's a bit different from what you mentioned.  It's question 1.

 Feb 2, 2016
 #3
avatar+128089 
+15
Best Answer

I'm confused, Yura_chan.....if   s(t)  =  sqrt(4t)   is the position function.....then....the average velocity  =  

 

[change in position]  / time traveled 

 

The  position  at t = 4 = sqrt(4*4)  = 4 cm

 

And the position at t = 1 = sqrt(4*1)  =  2cm

 

So ......

 

It moves 2cm  in 3 seconds    =  2/3  cm  per second....

 

 

cool cool cool

CPhill Feb 2, 2016
 #4
avatar+209 
+10

Sorry, you're correct. I just noticed the format they used in my textbook is the same that you mentioned.


Thank you!! :)

 Feb 2, 2016
 #5
avatar+118587 
+10

I wrote a big complement for you in here before Yura_chan and now it has disapeared.  There is not even a draft here for me to reaload sad

 

Anyway.....

 

I said thank you to Chris for answering your question.  wink

and

I said I was very pleased that you quized him when you didn't understand.

We want everyone to do that, it proves we really do have a student who is learning from us :)

 

You answer other people's questions great too.

 

We are very lucky that you have joined our web2.0calc forum   laughcoollaugh

 Feb 2, 2016
 #6
avatar+209 
+5

Thank you :) 

I love this site so much! Calc is really a k****r for me and there is no way I can pass the class without understanding the subject matter. All of you guys are super smart and helpful!! So thank you to Chris and you, and everyone else :)

 Feb 2, 2016

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