Answer:
Step-by-step explanation:
Given that acceleration of an object is
[tex]\frac{dv}{dt} =-2v\\\frac{dv}{v} =-2dt\\ln v = -2t+C\\[/tex]
is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence [tex]lnv = -2t+ln 7\\v=7e^{-2t}[/tex]
since velocity is rate of change of distance s we have
[tex]v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[[/tex]
substitute t=0 and s=0
[tex]C=7/2[/tex]
So solution for distance is
[tex]s(t) =\frac{-7}{2} (e^{-2t}-1)[/tex]