The position of a particle is given by s= f(t)=t^3-6t^2+9 where t is in seconds and s in meters ? a) when is the particle moving backward ( that is in the negative direction)? b) FInd the acceleration at time t and after 4s?

1 Answer
May 27, 2018

Position of a particle with respect to time is given as

s= f(t)=t^3-6t^2+9s=f(t)=t36t2+9

(a) Particle will move backwards when its velocity is negative -veve.

=>v(t)=dots=d/dt(t^3-6t^2+9)v(t)=.s=ddt(t36t2+9)
=>3t^2-12t3t212t

Imposing given condition we get

3t^2-12t<03t212t<0

Solutions is

0" > "t" < "4\s0 > t < 4s

(b) Acceleration a=dotva=.v

=>a(t)=d/dt(3t^2-12t)a(t)=ddt(3t212t)
=>a(t)=6t-12a(t)=6t12
Also a_4=6xx4-12=12\ ms^-2