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avatar+322 

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion 

\(s=\frac{4}{t^2}\),   where t is measured in seconds. Find the velocity of the particle at   t=a

 Feb 13, 2019
 #1
avatar+18965 
+2

The derivative of the displacement (or position) equation is the velocity function

   (then the NEXT deriviative is the acceleration function......just an FYI)

 

s = 4/t^2 = 4 t^-2

s ' = v = -8 t^-3             sub in t = a

        v at t= a  = -8 a^-3 = -8/a^3

 Feb 13, 2019
 #2
avatar+103131 
+1

Write  s   as   4t^(-2)

 

The velocity is the derivative  of the displacement function...so

 

s' (t)  =  -8t^(-3)

 

When t = a......the velocity is

 

s ' (a)  =  -8(a)^(-3)

 

 

cool cool cool

 Feb 13, 2019
 #3
avatar+322 
0

I'm confused on how you got the derivative.  I used f'(x)= [(x+h) - f(x)] / h   and I just have no clue what to do

Ruublrr  Feb 13, 2019
 #4
avatar+103131 
+1

OK....we have

 

[ f(t + h ) - f(t) ] / h  =

 

[ 4 / (t + h)^2   - 4/t^2 ] / h  =  

 

[  4t^2  - 4 ( t + h)^2 ] / [  h * t^2 * (t + h)^2 ]  =

 

[ 4t^2 - 4t^2 - 8ht - 4h^2 ] / [ h  * t^2 * ( t^2 + 2ht + h^2 ] =

 

[  h ( -8t - 4h) ] / [ h * ( t^4 + 2ht^3  + t^2h^2]  ....   [ divide out h ]

 

[ - 8t - 4h ] / [ (t^4 - 2ht^3 + t^2h^2 ]          let h →  0    and we have that

 

[ -8t ] /  [ t^4 ]  =

 

-8 / t^3     =  -8 t^(-3)

 

Don't worry....shortly.....you will learn to get to this result much faster  !!!!

 

 

cool cool cool

 Feb 13, 2019
edited by CPhill  Feb 13, 2019

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