Question

# The acceleration of a particle starting from rest vary with respect to time is given by ( a=(2 t-6) ), where tis in seconds. Find the time (in seconds) at which velocity of particle in negative direction is maximum.

Solution

( a=2 t-6 )

( frac{d v}{d t}=2 t-6 )

( d v=(2 r-6) d r )

( V=t^{2}-6 t )

( v rightarrow m a x )

( Rightarrow frac{d v}{d t}=0 )

( 2 t-6=0 )

( {t=3 operatorname{sec}} )