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)=t3−6t2+9
(a) Particle will move backwards when its velocity is negative
=>v(t)=dots=d/dt(t^3-6t^2+9)⇒v(t)=.s=ddt(t3−6t2+9)
=>3t^2-12t⇒3t2−12t
Imposing given condition we get
3t^2-12t<03t2−12t<0
Solutions is
0" > "t" < "4\s0 > t < 4s
(b) Acceleration
=>a(t)=d/dt(3t^2-12t)⇒a(t)=ddt(3t2−12t)
=>a(t)=6t-12⇒a(t)=6t−12
Alsoa_4=6xx4-12=12\ ms^-2