Question

A particle is moving with velocity v= t^3-6t^2+4 where v is in m/s and t is in seconds. At what time will the velocity be maximum/minimum and what is it equal to?

Answer 1

for max/min velocity..

d/dx(v) = 0

so, d/dx(t^3-6t^2+4)=0

3t^2-12t=0

3t-12=0

t=4sec

therfore at t=4sec the particle has max/min velocity..

hope u understood