We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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

Ruublrr Feb 13, 2019

#1**+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

ElectricPavlov Feb 13, 2019

#2**+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)

CPhill Feb 13, 2019

#4**+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 !!!!

CPhill Feb 13, 2019