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.

Find the time at which the particle switches from moving left to moving right.
t= ___ sec.

Answer 1

t=11 sec

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