The postion function of a particle is given by
s=t^3-1.5t^2-16t, t is greater than or equal to 0
A) when does the particle reach a velocity of 2m/s?
b) when is the acceleration 0?
s= t^3-1.5t^2-16t given as position function.
velocity is ds/dt = 3t^2-3t-16. we want velocity to be 2 m/s
so solve for 3t^2-3t-16=2
3t^2-3t =18 so t^2-t=6 then t^2-t-6=0 which factorises to
(t+2)(t-3)=0 giving a choice of solutions of t= -2 or t=3. we reject t=-2 as you can't have a negative answer here,so t=3,that is velocity is 2 m/s after 3seconds.
For acceleration,differentiate again ,ie we want to differentiate 3t^2-3t-16 this time because acceleration is the derivative of velocity. So now we get d/dt of 3t^2-3t-16 =6t-3 which we want to be zero.
6t-3=0 so the acceleration is zero after 0.5 seconds.
