a = -√v
v(dv/ds) = -√v
√vdv =-ds
2/3v3/2 = -s + c [integrating]
now write v as ds/dt and then integrate again
The accleration of a particle moving rectiliearly varies with the magnitude of its velocity as a = -√v
Find the displacement time equation.
i don't hav the answer...
so pls provide me with the answer....
a = -√v
v(dv/ds) = -√v
√vdv =-ds
2/3v3/2 = -s + c [integrating]
now write v as ds/dt and then integrate again