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?

Question

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?

anonymous · Answer

not 12 it should be 13m/s^2 ?

phi · Answer

I assume you found the 2nd derivative of  x  (which will be acceleration) ?

phi · Answer

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

phi · Answer

x'' = 12.82 m/s^2

anonymous · Answer

so 13 m/s^2 is correct .

phi · Answer

13 m/s^2 is correct if you round to the nearest whole number.