anonymous
  • anonymous
The position of an object is given by x= at^3 - bt^2 +ct , where a=4.1 m/s^3 , b=2.2 m/s^2, c= 1.7 m/s and x and t are in SI units. What is the instantaneuos acceleration of the object when t= 0.7 s ? I found 12 m/s^2 Am I correct?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
not 12 it should be 13m/s^2 ?
phi
  • phi
I assume you found the 2nd derivative of x (which will be acceleration) ?
phi
  • phi
x= at^3 - bt^2 +ct x' = 3at^2 -2bt + c x''= 6at - 2 b sub in for a, b, and t x'' = 6*4.1*0.7 - 2*2.2

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phi
  • phi
x'' = 12.82 m/s^2
anonymous
  • anonymous
so 13 m/s^2 is correct .
phi
  • phi
13 m/s^2 is correct if you round to the nearest whole number.

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