If the distance s described in time ' t ' by a particle moving on a straight line is given by s=t^5-40 t^3+30 t^2+80 t-250 , then its minimum acceleration is

Step-by-step solution

Given that, S=t^5-40 t^3+30 t^2+80 t-250 aligned & V= d S d t =5 t^4-120 t^2+60 t+80 & a= d V d t = d^2 S d t^2 =20 t^3-240 t+60 aligned Let aligned a & =f(t)=20 t^3-240 t+60 f^ (t) & =60 t^2-240 f^ (t) & =0 60 t^2-240=0 t^2=4 t & = 2 t=2, t=-2 f^ (t) & =120 t aligned at t=2, f^ (t)>0 So, at t=2, f(t) is minimum So, minimum acceleration is aligned a_ & =f(2)=20(2)^3-240 2+60 & =160-480+60=-260 . aligned