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A particle moves along the x-axis so that its velocity at any time t > 0 is given by v(t) = 4t(t – 7)(t – 1).

A. Find the acceleration at any time t.

 

B. Find the minimum acceleration of the particle over the interval [0, 3].

 

C. Find the maximum velocity of the particle over the interval [0, 2].

 Jan 13, 2020
 #1
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A.  

v(t)  = 4t(t - 7) (t - 1)     simplify

 

v(t)  = 4t  ( t^2 - 8t + 7)

 

v(t)  =  4t^3 - 32t^2 + 28t

 

The acceleration at any time  is the derivative of this  = v'(t)  = 12t^2 - 64t + 28

 

B.  Set the derivative to 0  and solve for t

 

12t^2 - 64t + 28  = 0        divide through by 4

 

3t^2 - 16t + 7     =  0         

 

The minimum acceleration on [0,3 ] occurs at  t  =    (16)/ (2 * 3)  =  16/6 =  8/3  sec

 

And the minimum acceleration is   3(8/3)^2  - 16(8/3) + 7   =  -43/3

 

C. See the graph here :  https://www.desmos.com/calculator/70cdiunnem

 

The max velocity on [0, 2 ] occurs at about .481 s   and is  ≈ 6.51

 

 

cool cool cool 

 Jan 13, 2020

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