Question

(1. suppose the postition of an object moving horizontally after t seconds is given by the function s = f(t) = 2t^3 - 21 t^2 + 60 t where s is measured in feet, with s > 0 corresponding to positions right of the orgin. EXPLAIN ALL THE STEPS!

A) Find the velocity and acceleration of the object.

B) Determine the acceleration of the object when its velocity is zero.

Answer 1

@AccessDenied @zepdrix @robtobey

Answer 2

The velocity function is given by the derivative of the position function s.\[\Large\sf s'(t)=v(t)\]So our s(t) is,\[\Large\sf s(t)=2t^3-21t^2+60t\]So what is the derivative?\[\Large\sf s'(t)=?\]Apply the power rule to each term.

Answer 3

Can you please show me how to apply the power rule and step by step so I can understand?

Answer 4

@zepdrix

Answer 5

@ayeshaafzal221 @ChristinaWiler @dumbsearch2 @elaornelas @eliassaab @ifrimpanainte @Kira_Yamato @kirbykirby

Answer 6

@nitz

Answer 7

The power rule is:\[d/dx(x ^{n}) = nx ^{n-1}\]

Answer 8

The another result which will be used over here is that derivative of sum or difference of two functions is equal to sum or difference of their independent derivatives wrt to the variable..

Answer 9

d/dx ((f(x)+g(x)) = d/dx(f(x)) + d/dx((g(x))

Answer 10

Always remember if y=f(x) ,then the derivative is dy/dx
ie. derivative of dependent wrt to independent variable

Answer 11

dy/dx or \[f \prime (x)\] means the same thing...derivative of y wrt x

Answer 12

So,\[s \prime (t) =6t ^{2}-42t +60\]

Answer 13

What would the acceleration be for B?

Answer 14

acceleration is \[s \prime \prime (t) \]

Answer 15

So it is s'(t) = 6t^3 - 42t + 60 that's for B?

Answer 16

@hartnn

Answer 17

@washcaps @pgpilot326 @praxer @Miamy

Answer 18

@RadEn @Rvc @richyw

Answer 19

no, its 12t - 42

Answer 20

Here is a complete solution
\[
s(t)=2 t^3-21 t^2+60 t\\
s'(t)=6 t^2-42 t+60\\
s''(t)=12 t-42\\
s'(t) =0 \text { for } t=2, t=5\\
s''(2)=-18\\
s''(5)=18\\

\]