Question

A particle is moving horizontally along the x-axis. Its position (in ft) is: s(t)=t^3-18t^2+33t+14 where t is in sec.

Answer 1

Answer:

t=11 sec

Step-by-step explanation:

The position of the particle moving along the x-axis is given by:

$s(t) = {t}^{3} - 18 {t}^{2} + 33t + 14$

The velocity is given by:

$s'(t) = 3 {t}^{2} - 36{t} + 33$

If s'(t)>0 then the particle is moving right.

$3 {t}^{2} - 36{t} + 33 \: > \: 0 \\ {t}^{2} - 12{t} + 11\: > \: 0$

$\implies \: t \: < \: 1 \: or \: t \: > \: 11$

This means that the particle is moving left when

$1 \: < \: t \: < \: 11$

The particle changes direction at time t=1 or t=11