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avatar+956 

Let f(t) = 2t3 represent the position of a particle in feet at time t in seconds on the interval 0 ≤ t ≤ 8.

 May 14, 2020
 #1
avatar+111438 
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I'll  do  the first one  in (a) .....you should be able to  do the rest based on this one, GM

 

Average  velocity   =    [ 2(5)^3  - 2(2)^3 ] / [ 5 - 2]    =  [ 150 - 16 ] / 3   ≈  44.66  ft /s

 

 

For the second one....we   can use  the   result  of  the  V part of a

 

[ 2(5.001)^3  - 2(5)^3 ]  /  [ 5.001  - 5 ]   ≈  150 ft/s

 

{ BTW.....Calculus  can show that this is true.....the instantaneous velocity at 5 sec  =  150 ft / s   exactly }

 

cool cool cool

 May 14, 2020

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