The acceleration of a particle sta...
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.

JEE/Engineering Exams
Physics
Solution
110
5.0 (1 ratings)

( 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}} )