Question

( a d t=0, quad v=0 )
[
a=sin pi t+cos pi z
]
( frac{d V}{d t}=sin pi t+cos pi t )
( Rightarrow int d V=intleft(sin ^{pi} t+cos pi tright) d t )
( Rightarrow quad v=frac{sin pi t-cos pi t}{pi}+c )
ad ( z=0, quad v=0 )
( Rightarrow 0=-frac{1}{pi}+c Rightarrow c=frac{1}{pi} )
( therefore quad V(t)=frac{sin pi z-cos pi t+1}{pi} )
( V_{max } ) when ( frac{d V}{d t}=0 Rightarrow a=0 )
( sin pi t+cos pi t=0 )
2) tantt =
( pi t=frac{3 pi}{4} )
2) ( quad t=23 / 4 )
( m a x=sin 3 frac{pi}{4}-c d )
( =frac{frac{1}{2}+frac{1}{2}}{pi}+1=frac{sqrt{2}+}{pi} )

# A particle is at rest at t = 0. If acceleration of the particle is given as a = sinnt + cosnt, in SI units, then the maximum speed of particle is 212 mis (1) = m/s (3) (V2+1) mis (4) ANZ mis TT 09456

Solution