Question

A particle starts at the origin and moves along the positive x-axes such that its distance from the origin at any time is determined by the equation: x=2t^3 -9t^2 +12t Find the 3.1) times when the velocity is zero. 3.2) acceleration when the velocity is zero.

Answer 1

3.1) times when the velocity is zero.

we are given the equation $X=2t^3 -9t^2 +12t$

we need to differentiate this function with respect to t to obtain velocity equation

$\implies$ ${dX\over dt}=V(t)=6t^2-18t+12$

we are asked to find times when velocity is zero

$\implies$ $V(t)=6t^2-18t+12=0$

divide by 6 through

${6t^2\over 6}-{18t\over 6}+{12\over 6}=0$

$\implies$ $(t-1)(t-2)=0$

$\therefore t=1$ sec and $t=2$ secs

3.2) acceleration when the velocity is zero.

To get acceleration, we differentiate velocity equation with respect to time

$\implies{dV\over dt}=a(t)=12t-18$

when $t=2$

$\implies a(2)=12(2)-18=6$ $m/s^2$

When $t=1$

$\implies a(1)=12(1)-18=-6$ $m/s^2$

$\therefore$ decelarition when t=1 is 6 $m/s^2$