Answer:
47.91 sec
Explanation:
it is given that [tex]\alpha =\frac{1}{4v^{2}}[/tex]
at t=0 velocity =0 ( as it is given that it is starting from rest )
we have to find time at which velocity will be 3.3 [tex]\frac{m}{sec^{2}}[/tex]
we know that [tex]\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}[/tex]
[tex]4v^{2}dv=dt[/tex]
integrating both side
[tex]\frac{4v^{3}}{3}=t+c[/tex]---------------eqn 1
at t=o it is given that v=0 putting these value in eqn 1 c=0
so [tex]\frac{4v^{3}}{3}=t[/tex]
when v= 3.3 [tex]\frac{m}{sec^{2}}[/tex]
t=[tex]\frac{4}{3}\times 3.3^{3}[/tex]
=47.91 sec