Question

The position of an object as a function of time is given by `x(t)= at^3-bt^2+ct-d` where a=3.6 m/s^2, b=4.0 m/s^2, c=60m, and d=7.0m, how do you find instantaneous acceleration at t=2.4s?

Answer 1

Given,

`x=at^3 -bt^2 +ct-d`

So, velocity =rate of change in displacement=`(dx)/(dt) =v=3at^2 -2bt+c`

Again,accleration = rate of change in velocity=`(dv)/(dt)=6at-2b`

Given, `a=3.6,b=4`

so, `|(dv)/(dt)|_(t=2.4)=6*3.6*2.4 - 2*4=43.84ms^-2`