Question

[
begin{aligned}
u_{1}=& 0, a_{1}=2 m_{s}^{2}
v_{1}=& u_{1}+a_{1} t_{1}=0+2left(t_{1}right)=2 t_{1}
end{aligned}
]
When it staits to setaid.
[
begin{array}{l}
u_{2}=v_{2}=2 t_{1}
v_{2}=0, quad a=-sin left(s^{2}right.
end{array}
]
( therefore quad V_{2}=u_{2}+a_{2} t_{2} )
& ( D=2 t_{1}+(-3)left(t_{2}right) )
( x_{t}=3 t_{2} )
and ( quad t_{1}+t_{2}=10 )
( therefore t_{1}=b s )
th
A ( quad V_{text {inct }}=V_{1}=2 t_{1}=2(b)=ln M / s )

# A body starts from rest with an acceleration 2 m/s2 till it attains the maximum velocity then retards to rest with 3 m/s2. If total time taken is 10 second, then maximum speed attained is

Solution