Question

the position of a moving particle is given s(t)=2t^3-21t^2+60t where distance is measured in ft & time in secs. find velocity of particle at t=1

Answer 1

s(t)=2t^3-21t^2+60t
v(t)=d(s(t))/dt (v=velocity)
=6t^2-42t+60

by puting the value of t we get
v=6-42+60 =24 ft/sec