+0  
 
0
98
7
avatar+138 

So I took the derivate of V to get the acceleration, but it ends up with displacement still in the equation I believe, so how do I do this.

 Jan 31, 2021
 #1
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The acceleration and velocity are dependent on the displacement ....so the displacement still needs to be in the equation for accel.

 Jan 31, 2021
 #2
avatar+138 
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okay well how do I solve displacement? The answer for a is a single numerical value, without variables, so I have to be able to solve it somehow.

SpaceTsunaml  Jan 31, 2021
 #3
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You are given the displacement of   s = 3

    put that in your accel equation to calc accel

Guest Jan 31, 2021
 #4
avatar+138 
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im talking about question a. the way these problems are supposed to work is a is seperate from b, but b relys on a

SpaceTsunaml  Jan 31, 2021
 #5
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This question a) and b) are not dependent....

 

maximum velocity ...and acceleration at s = 3    are not the same

    take derivative of velocity , set = 0    to find s value where velocity is maximum

          use this value of 's' in the given velocity equation to calculate the VALUE of the velocity at the 's' you found....

Guest Jan 31, 2021
 #6
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Here is a graph that may help you verify the answers you find:

 

 Jan 31, 2021
 #7
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Don't know where ya went....

   derivative of the velocity function is the acceleration function....you got that !

      derivative of the velocity function also shows the slope of the velocity function at each point

             where the slope goes from positive to negative    or   from  negative to positive  

                  the acceleation is too....these are the points where velocity will be a maximum (maybe negative or  maybe positive value)

                      these occur where the acceleration function = 0   (slope = 0    as it goes from pos to neg or  neg to pos)

 

 

the derivative of the given velocity funtion is given as    -10(s^2-4s-4) / (s+4)^2

                                                                                       -10(s^2-4s-4)/(s+4)^2 = 0

                                                                                         -10s^2 +40s+40 = 0                 Use Quadratic Formula

                                                                                             s = 2+- 2 sqrt2               these are the points where accel = 0

 

 

Plug these values into the orignal Velocity equation to calculate the values......pick the one that is maximum (here I do not know if they want most positive or greatest magnitude)

 

Use s =3    in the derived ACCELERATION equation to calculate the acceleration at s = 3 

 

   The graph above should help to verify your answers......  

 Jan 31, 2021

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